A computer retail store has 1414 personal computers in stock. A buyer wants to purchase 33 of them. Unknown to either the retail store or the buyer, 33 of the computers in stock have defective hard drives. Assume that the computers are selected at random.
A) In how many different ways can the 3 computers be chosen? 120
B) What is the probability that exactly one of thecomputers will be defective?
C) What is the probability that at least one of thecomputers selected is defective?

Answers

Answer 1

Answer:

a) 364 ways

b) 45.33% probability that exactly one of the computers will be defective.

c) 54.67% probability that at least one of the computers selected is defective.

Step-by-step explanation:

The computers are chosen without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this question:

14 computers, so N = 14.

3 defective, so k = 3.

3 will be purchesed, so n = 3.

A) In how many different ways can the 3 computers be chosen?

3 from a set of 14. So

[tex]C_{14,3} = \frac{14!}{3!(14-3)!} = 364[/tex]

364 ways

B) What is the probability that exactly one of thecomputers will be defective?

This is P(X = 1).

[tex]P(X = 1) = h(1,14,3,3) = \frac{C_{3,1}*C_{11,2}}{C_{14,3}} = 0.4533[/tex]

45.33% probability that exactly one of the computers will be defective.

C) What is the probability that at least one of the computers selected is defective?

Either none is, or at least one is. The sum of the probabilities of these events is 1. So

[tex]P(X = 0) + P(X \geq 1) = 1[/tex]

We want [tex]P(X \geq 1)[/tex]. Then

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which

[tex]P(X = 0) = h(0,14,3,3) = \frac{C_{3,0}*C_{11,3}}{C_{14,3}} = 0.4533[/tex]

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.4533 = 0.5467[/tex]

54.67% probability that at least one of the computers selected is defective.


Related Questions

Which expression is equivalent to pq

Answers

Answer:

D

Step-by-step explanation:

Mark Brainliest

The life of an electric component has an exponential distribution with a mean of 8.9 years. What is the probability that a randomly selected one such component has a life more than 8 years? Answer: (Round to 4 decimal places.)

Answers

Answer:

[tex] P(X>8)[/tex]

And for this case we can use the cumulative distribution function given by:

[tex] F(x) = 1- e^{-\lambda x}[/tex]

And if we use this formula we got:

[tex] P(X>8)= 1- P(X \leq 8) = 1-F(8) = 1- (1- e^{-\frac{1}{8.9} *8})=e^{-\frac{1}{8.9} *8}= 0.4070[/tex]

Step-by-step explanation:

For this case we can define the random variable of interest as: "The life of an electric component " and we know the distribution for X given by:

[tex]X \sim exp (\lambda =\frac{1}{8.9}) [/tex]

And we want to find the following probability:

[tex] P(X>8)[/tex]

And for this case we can use the cumulative distribution function given by:

[tex] F(x) = 1- e^{-\lambda x}[/tex]

And if we use this formula we got:

[tex] P(X>8)= 1- P(X \leq 8) = 1-F(8) = 1- (1- e^{-\frac{1}{8.9} *8})=e^{-\frac{1}{8.9} *8}= 0.4070[/tex]

pqrs is a rhombus. If PO= 4 cm and OQ=3cm,then find PQ.​(please answer fast)

Answers

Answer: 5 cm

Step-by-step explanation:

The diagonals of a rhombus bisect each other in half

PO = OR = 4 cm

So. PR = 8 cm

Similarly,

SO = OQ = 3 cm

So, SQ = 6 cm

To measure side, formula is

a = √p^2 + q^2 / 2

a = √6^2 + 8^2 / 2

a = √36+64 / 2

a = √100 / 2

a = 10/2 = 5 cm

When planning a more strenuous hike, Nadine figures that she will need at least 0.6 liters of water for each hour on the trail. She also plans to always have at least 1.25 liters of water as a general reserve. If x represents the duration of the hike (in hours) and y represents the amount of water needed (in liters) for a hike, the following inequality describes this relation: y greater or equal than 0.6 x plus 1.25 Which of the following would be a solution to this situation?

Answers

Answer:

The solution for this is:

y = (0.6 * x) + 1.25

Hope it helps! :)

Answer:

Having 3.2 liters of water for 3 hours of hiking

Step-by-step explanation:

If x represents the number of hours and y represents the number of liters of water, then we can plug the possible solutions into our inequality to see which solution(s) work.

The first option is having 3 liters of water for 3.5 hours of hiking. We will plug 3 in for y and 3.5 in for x:

y > 0.6x + 1.25

3 > 0.6(3.5) + 1.25

3 > 3.35

But since 3 is not greater than 3.35, this does not work.

The next option is having 2 liters of water for 2.5 hours of hiking:

2 > 0.6(2.5) + 1.25

2 > 2.75

But 2 is not greater than 2.75, so this does not work.

Option c is having 2.3 liters of water for 2 hours of hiking:

2.3 > 0.6(2) + 1.25

2.3 > 2.45

Since 2.3 is not greater than 2.45, this solution does not work.

The last option is having 3.2 liters of water for 3 hours of hiking:

3.2 > 0.6(3) + 1.25

3.2 > 3.05

3.2 IS greater than 3.05, so this solution works!

Brian invests £8000 into his bank account. He receives 3% per year compound interest. How many years will it take for Brian to have more than £9500?

Answers

Answer:

6 years is the correct answer.

Step-by-step explanation:

Given that

Principal, P =  £8000

Rate of interest, R = 3% compounding annually

Amount, A >  £9500

To find: Time, T = ?

We know that formula for Amount when interest in compounding:

[tex]A = P \times (1+\dfrac{R}{100})^T[/tex]

Putting all the values:

[tex]A = 8000 \times (1+\dfrac{3}{100})^T[/tex]

As per question statement, A >  £9500

[tex]\Rightarrow 8000 \times (1+\dfrac{3}{100})^T > 9500\\\Rightarrow (1+0.03)^T > \dfrac{9500}{8000}\\\Rightarrow (1.03)^T > 1.19[/tex]

Putting values of T, we find that at T = 6

[tex]1.03^6 = 1.194 > 1.19[/tex]

[tex]\therefore[/tex] Correct answer is T = 6 years

In 6 years, the amount will be more than £9500.

Solve for X. Show all work

Answers

Answer:

About 11.77 centimeters

Step-by-step explanation:

By law of sines:

[tex]\dfrac{50}{\sin 62}=\dfrac{x}{\sin 12} \\\\\\x=\dfrac{50}{\sin 62}\cdot \sin 12\approx 11.77cm[/tex]

Hope this helps!

. A bag contains 6 red and 3 black chips. One chip is selected, its color is recorded, and it is returned to the bag. This process is repeated until 5 chips have been selected. What is the probability that one red chip was selected?

Answers

Answer:

The probability that one red chip was selected is 0.0053.

Step-by-step explanation:

Let the random variable X be defined as the number of red chips selected.

It is provided that the selections of the n = 5 chips are done with replacement.

This implies that the probability of selecting a red chip remains same for each trial, i.e. p = 6/9 = 2/3.

The color of the chip selected at nth draw is independent of the other selections.

The random variable X thus follows a binomial distribution with parameters n = 5 and p = 2/3.

The probability mass function of X is:

[tex]P(X=x)={5\choose x}\ (\frac{2}{3})^{x}\ (1-\frac{2}{3})^{5-x};\ x=0,1,2...[/tex]

Compute the probability that one red chip was selected as follows:

[tex]P(X=1)={5\choose 1}\ (\frac{2}{3})^{1}\ (1-\frac{2}{3})^{5-1}[/tex]

                [tex]=5\times\frac{2}{3}\times \frac{1}{625}\\\\=\farc{2}{375}\\\\=0.00533\\\\\approx 0.0053[/tex]

Thus, the probability that one red chip was selected is 0.0053.

Answer:

0.0412

Step-by-step explanation:

Total chips = 6 red + 3 black chips

Total chips=9

n=5

Probability of (Red chips ) can be determined by

=[tex]\frac{6}{9}[/tex]

=[tex]\frac{2}{3}[/tex]

=0.667

Now we used the binomial theorem

[tex]P(x) = C(n,x)*px*(1-p)(n-x).....Eq(1)\\ putting \ the \ given\ value \ in\ Eq(1)\ we \ get \\p(x=1) = C(5,1) * 0.667^1 * (1-0.667)^4[/tex]

This can give 0.0412

please - i got this wrong so plz help

Answers

Answer:

Area = 108 cm^2

Perimeter = 44 cm

Step-by-step explanation:

Area, -->

24 + 30 + 24 + 30 -->

24(2) + 30(2)

48 + 60 = 108 cm^2

108 = area

10 + 12 + 10 + 12, -->

10(2) + 12(2) = 44 cm

44 = perim.

Hope this helps!

Answer:

Step-by-step explanation:

Draw the diagram.

This time put in the only one line for the height. That is only 1 height is 8 cm. That's it.

The base is 6 +  6 = 12 cm.

The slanted line is 10 cm

That's all your diagram should show. It is much clearer without all the clutter.

Now you are ready to do the calculations.

Area

The Area = the base * height.

base = 12

height = 8

Area = 12 * 8 = 96

Perimeter.

In a parallelagram the opposite sides are equal to one another.

One set of sides = 10 + 10 = 20

The other set = 12 + 12 = 24

Both sets = 20 + 24

Both sets = 44

Answer

Area = 96

Perimeter = 44

A​ student's course grade is based on one midterm that counts as 20​% of his final​ grade, one class project that counts as 20​% of his final​ grade, a set of homework assignments that counts as 30​% of his final​ grade, and a final exam that counts as 30​% of his final grade. His midterm score is 64​, his project score is 80​, his homework score is 94​, and his final exam score is 77. What is his overall final​ score? What letter grade did he earn​ (A, B,​ C, D, or​ F)? Assume that a mean of 90 or above is an​ A, a mean of at least 80 but less than 90 is a​ B, and so on.

Answers

Answer:

His overall final​ score is 80.1.

His letter grade is a B.

Step-by-step explanation:

To find his grade, we multiply each grade by it's weight.

Grades and weights:

His midterm score is 64​. The midterm counts 20% = 0.2.

His project score is 80​. The project score counts 20% = 0.2.

His homework score is 94. The homework score counts 30%.

His final exam score is 77. It counts 30%.

What is his overall final​ score?

64*0.2 + 80*0.2 + 94*0.3 + 77*0.3 = 80.1

His overall final​ score is 80.1.

What letter grade did he earn​ (A, B,​ C, D, or​ F)?

At least 80 but less than 90 is a​ B. He scored 80.1, so his letter grade is a B.

Find an equation for this line.

Answers

Answer:

y = -0.4x - 3

Step-by-step explanation:

Using the slope formula, y2-y1/x2-x1 we need to find two points. Luckily, we already have two points, (5, -5) and (-5, -1). Plugging in, we have -4/10, or -0.4. Since now we know m = -0.4, we need to find the y-intercept. We have it as -3. Now we get y = -0.4x - 3 as our equation.

I. In the testing of a new production method, 18 employees were selected randomly and asked to try the new method. The sample mean production rate for the 18 employees was 80 parts per hour and the sample standard deviation was 10 parts per hour. Provide 90% confidence intervals for the populations mean production rate for the new method, assuming the population has a normal probability distribution.

Answers

Answer:

The 90% confidence interval for the mean production rate fro the new method is (75.9, 84.1).

Step-by-step explanation:

We have to calculate a 90% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=80.

The sample size is N=18.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{10}{\sqrt{18}}=\dfrac{10}{4.24}=2.36[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=18-1=17[/tex]

The t-value for a 90% confidence interval and 17 degrees of freedom is t=1.74.

The margin of error (MOE) can be calculated as:

[tex]MOE=t\cdot s_M=1.74 \cdot 2.36=4.1[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL=M-t \cdot s_M = 80-4.1=75.9\\\\UL=M+t \cdot s_M = 80+4.1=84.1[/tex]

The 90% confidence interval for the mean production rate fro the new method is (75.9, 84.1).

A manager bought 12 pounds of peanuts for $30. He wants to mix $5 per pound cashews with the peanuts to get a batch of mixed nuts that is worth $4 per pound. How many pounds of cashews are needed

Answers

Answer:

18 pounds of cashews are needed.

Step-by-step explanation:

Given;

A manager bought 12 pounds of peanuts for $30.

Price of peanut per pound P = $30/12 = $2.5

Price of cashew per pound C = $5

Price of mixed nut per pound M = $4

Let x represent the proportion of peanut in the mixed nut.

The proportion of cashew will then be y = (1-x), so;

xP + (1-x)C = M

Substituting the values;

x(2.5) + (1-x)5 = 4

2.5x + 5 -5x = 4

2.5x - 5x = 4 -5

-2.5x = -1

x = 1/2.5 = 0.4

Proportion of cashew is;

y = 1-x = 1-0.4 = 0.6

For 12 pounds of peanut the corresponding pounds of cashew needed is;

A = 12/x × y

A = 12/0.4 × 0.6 = 18 pounds

18 pounds of cashews are needed.

According to a report an average person watched 4.55 hours of television per day in 2005. A random sample of 20 people gave the following number of hours of television watched per day for last year. At the 10% significance level, do the data provide sufficient evidence to conclude that the amount of television watched per day last year by the average person differed from that in 2005? 1.0 4.6 5.4 3.7 5.2 1.7 6.1 1.9 7.6 9.1 6.9 5.5 9.0 3.9 2.5 2.4 4.7 4.1 3.7 6.2 a. identify the claim and state and b. find the critical value(s) and identify the rejection region(s), c. find the standardized test statistic Sketch a graph decide whether to reject or fail to reject the null hypothesis, and d. interpret the decision in the context of the original claim. e. Obtain a 95%confidence interval

Answers

Answer:

a. The claim is that the amount of television watched per day last year by the average person differed from that in 2005.

b. The critical values are tc=-1.729 and tc=1.729.

The acceptance region is defined by -1.792<t<1.729. See the picture attached.

c. Test statistic t=0.18.

The null hypothesis failed to be rejected.

d. At a significance level of 10%, there is not enough evidence to support the claim that the amount of television watched per day last year by the average person differed from that in 2005.

e. The 95% confidence interval for the mean is (2.29, 7.23).

Step-by-step explanation:

We have a sample of size n=20, which has mean of 4.76 and standard deviation of 5.28.

[tex]M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{20}(1+4.6+5.4+. . .+6.2)\\\\\\M=\dfrac{95.2}{20}\\\\\\M=4.76\\\\\\s=\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2\\\\\\s=\dfrac{1}{19}((1-4.76)^2+(4.6-4.76)^2+(5.4-4.76)^2+. . . +(6.2-4.76)^2)\\\\\\s=\dfrac{100.29}{19}\\\\\\s=5.28\\\\\\[/tex]

a. This is a hypothesis test for the population mean.

The claim is that the amount of television watched per day last year by the average person differed from that in 2005.

Then, the null and alternative hypothesis are:

[tex]H_0: \mu=4.55\\\\H_a:\mu\neq 4.55[/tex]

The significance level is 0.1.

The sample has a size n=20.

The sample mean is M=4.76.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=5.28.

The estimated standard error of the mean is computed using the formula:

[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{5.28}{\sqrt{20}}=1.181[/tex]

Then, we can calculate the t-statistic as:

[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{4.76-4.55}{1.181}=\dfrac{0.21}{1.181}=0.18[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=20-1=19[/tex]

The critical value for a level of significance is α=0.10, a two tailed test and 19 degrees of freedom is tc=1.729.

The decision rule is that if the test statistic is above tc=1.729 or below tc=-1.729, the null hypothesis is rejected.

As the test statistic t=0.18 is within the critical values and lies in the acceptance region, the null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the amount of television watched per day last year by the average person differed from that in 2005.

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=4.76.

The sample size is N=20.

The standard error is s_M=1.181

The degrees of freedom for this sample size are df=19.

The t-value for a 95% confidence interval and 19 degrees of freedom is t=2.093.

The margin of error (MOE) can be calculated as:

[tex]MOE=t\cdot s_M=2.093 \cdot 1.181=2.47[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL=M-t \cdot s_M = 4.76-2.47=2.29\\\\UL=M+t \cdot s_M = 4.76+2.47=7.23[/tex]

The 95% confidence interval for the mean is (2.29, 7.23).

What is the simplified form of the expression 3cubed root b^2

Answers

Answer:

Step-by-step explanation:

[tex](\sqrt{b^{2}})^{3}=b^{3}\\\\[/tex]

or If it is

[tex]\sqrt[3]{b^{2}} =(b^{2})^{\frac{1}{3}}=b^{2*\frac{1}{3}}=b^{\frac{2}{3}}[/tex]

Which is equivalent to V180x11 after it has been simplified completely?



© 3x10653



O 3x® v5x



O 6x10/5x



O 6x55x

Answers

Question:

Which is equivalent to [tex]\sqrt{180x^{11}}[/tex] after it has been simplified completely?

Answer:

[tex]\sqrt{180x^{11}} = 6x^{5}\sqrt{5x}[/tex]

Step-by-step explanation:

Given

[tex]\sqrt{180x^{11}}[/tex]

Required

Simplify

We start by splitting the square root

[tex]\sqrt{180x^{11}} = \sqrt{180} * \sqrt{x^{11}}[/tex]

Replace 180 with 36 * 5

[tex]\sqrt{180x^{11}} = \sqrt{36 * 5} * \sqrt{x^{11}}[/tex]

Further split the square roots

[tex]\sqrt{180x^{11}} = \sqrt{36} *\sqrt{5} * \sqrt{x^{11}}[/tex]

[tex]\sqrt{180x^{11}} = 6*\sqrt{5} * \sqrt{x^{11}}[/tex]

Replace power of x; 11 with 10 + 1

[tex]\sqrt{180x^{11}} = 6*\sqrt{5} * \sqrt{x^{10 + 1}}[/tex]

From laws of indices; [tex]a^{m+n} = a^m * a^n[/tex]

So, we have

[tex]\sqrt{180x^{11}} = 6*\sqrt{5} * \sqrt{x^{10} * x^1}[/tex]

[tex]\sqrt{180x^{11}} = 6*\sqrt{5} * \sqrt{x^{10} * x}[/tex]

Further split the square roots

[tex]\sqrt{180x^{11}} = 6*\sqrt{5} * \sqrt{x^{10}} * \sqrt{x}[/tex]

From laws of indices; [tex]\sqrt{a} = a^{\frac{1}{2}}[/tex]

So, we have

[tex]\sqrt{180x^{11}} = 6*\sqrt{5} * x^{10*\frac{1}{2}} * \sqrt{x}[/tex]

[tex]\sqrt{180x^{11}} = 6*\sqrt{5} * x^{\frac{10}{2}} * \sqrt{x}[/tex]

[tex]\sqrt{180x^{11}} = 6*\sqrt{5} * x^{5} * \sqrt{x}[/tex]

Rearrange Expression

[tex]\sqrt{180x^{11}} = 6 * x^{5} * \sqrt{5} * \sqrt{x}[/tex]

[tex]\sqrt{180x^{11}} = 6x^{5} * \sqrt{5} * \sqrt{x}[/tex]

From laws of indices; [tex]\sqrt{a} *\sqrt{b} = \sqrt{a*b} = \sqrt{ab}[/tex]

So, we have

[tex]\sqrt{180x^{11}} = 6x^{5} * \sqrt{5*x}[/tex]

[tex]\sqrt{180x^{11}} = 6x^{5} * \sqrt{5x}[/tex]

[tex]\sqrt{180x^{11}} = 6x^{5}\sqrt{5x}[/tex]

The expression can no longer be simplified

Hence, [tex]\sqrt{180x^{11}}[/tex] is equivalent to [tex]6x^{5}\sqrt{5x}[/tex]

Answer:

D on edg

Step-by-step explanation:

6x^5 V5x is correct

The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal​ distribution, with a mean of 17 minutes and a standard deviation of 4 minutes. ​(a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take​ longer, the customer will receive the service for​ half-price. What percent of customers receive the service for​ half-price? ​(b) If the automotive center does not want to give the discount to more than 2​% of its​ customers, how long should it make the guaranteed time​ limit?

Answers

Answer:

a) 22.66% of customers receive the service for​ half-price.

b) The guaranteed time​ limit should be of 25.2 minutes.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

[tex]\mu = 17, \sigma = 4[/tex]

​(a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take​ longer, the customer will receive the service for​ half-price. What percent of customers receive the service for​ half-price?

Longer than 20 minutes is 1 subtracted by the pvalue of Z when X = 20. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{20 - 17}{4}[/tex]

[tex]Z = 0.75[/tex]

[tex]Z = 0.75[/tex] has a pvalue of 0.7734

1 - 0.7734 = 0.2266

22.66% of customers receive the service for​ half-price.

(b) If the automotive center does not want to give the discount to more than 2​% of its​ customers, how long should it make the guaranteed time​ limit?

The time limit should be the 100 - 2 = 98th percentile, which is X when Z has a pvalue of 0.98. So X when Z = 2.054.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]2.054 = \frac{X - 17}{4}[/tex]

[tex]X - 17 = 4*2.054[/tex]

[tex]X = 25.2[/tex]

The guaranteed time​ limit should be of 25.2 minutes.

What is the area of this triangle?

Answers

Answer:

Option (D)

Step-by-step explanation:

Formula for the area of a triangle is,

Area of a triangle = [tex]\frac{1}{2}(\text{Base})(\text{Height})[/tex]

For the given triangle ABC,

Area of ΔABC = [tex]\frac{1}{2}(\text{AB})(\text{CD})[/tex]

Length of AB = [tex](y_2-y_1)[/tex]

Length of CD = [tex](x_3-x_1)[/tex]

Now area of the triangle ABC = [tex]\frac{1}{2}(y_2-y_1)(x_3-x_1)[/tex]

Therefore, Option (D) will be the answer.

Create a set of data that shows temperature highs for 10 days and satisfies each condition below:


Mean: 72

Median: 74

Mode: 68

Range: 21

Answers

72*10=720 so all the numbers would need to add to 720

the median is 74 so you need to have both 75 and 76 in the set

the mode is 68 so that need to be in at least twice

and the range is 21 so the largest number-21=smallest number

57, 68, 68, 68, 75, 76, 76, 77, 77, 78

A set of data that shows temperature highs for 10 days is 57, 68, 68, 68, 75, 76, 76, 77, 77, and 78.

Given that, create a set of data that shows temperature highs for 10 days.

What are the Mean Median and Mode?

Mean, median and mode are all measures of central tendency in statistics. In different ways, they each tell us what value in a data set is typical or representative of the data set.

The mean is the same as the average value of a data set and is found using a calculation. Add up all of the numbers and divide by the number of numbers in the data set.

The median is the central number of a data set. Arrange data points from smallest to largest and locate the central number. This is the median. If there are 2 numbers in the middle, the median is the average of those 2 numbers.

The mode is the number in a data set that occurs most frequently. Count how many times each number occurs in the data set. The mode is the number with the highest tally. It's ok if there is more than one mode. And if all numbers occur the same number of times there is no mode.

Now,

72×10=720 so all the numbers would need to add to 720.

The median is 74 so you need to have both 75 and 76 in the set.

The model is 68 so that needs to be in at least twice.

The range is 21 so the largest number-21=smallest number

57, 68, 68, 68, 75, 76, 76, 77, 77, 78

Therefore, a set of data that shows temperature highs for 10 days is 57, 68, 68, 68, 75, 76, 76, 77, 77, and 78.

To learn more about the Mean Median and Mode visit:

https://brainly.com/question/3183994.

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The translation (x − 10, y + 17) is applied to a triangle. Maryanne makes a conjecture about the perimeter of the image of the triangle, tests the conjecture, and finds that it is true. What could have been her conjecture?

Answers

Answer:

See below.

Step-by-step explanation:

Translations do not change the perimeter (nor the area for that matter). Therefore, her conjecture could be that: "After translating this triangle 10 units to the left and 17 units upwards, the perimeter will be the same."

A randomly selected sample of college basketball players has the following heights in inches. See Attached Excel for Data. Compute a 95% confidence interval for the population mean height of college basketball players based on this sample and fill in the blanks appropriately. < μ < (round to 3 decimal places)

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  confidence interval is  [tex]64.86<\mu<67[/tex]

Step-by-step explanation:

From the question we are given the following data

   The following heights are

66, 65, 67, 62, 62, 65, 61, 70, 66, 66, 71, 63, 69, 65, 71, 66, 66, 69, 68, 62, 65, 67, 65, 71, 65, 70, 62, 62, 63, 64, 67, 67      

 The  sample size is n  =32

  The confidence level is [tex]k = 95[/tex]% = 0.95

The mean is evaluated as

          [tex]\= x = 66+ 65+ 67+ 62+ 62+ 65+ 61+ 70+ 66+ 66+ 71+63+ 69+ 65+ 71+ 66+ 66+ 69+ 68+ 62+ 65+ 67,+\\65+ 71+ 65+ 70+ 62+ 62+ 63+ 64+ 67+ 67 / 32[/tex]

=>   [tex]\= x = \frac{2108}{32}[/tex]

=>     [tex]\= x = 65.875[/tex]

The standard deviation is evaluated as

           [tex]\sigma = \sqrt{ v}[/tex]

Now  

   [tex]v = ( 66-65.875 )^2+(65-65.875)^2+( 67-65.875)^2+ (62-65.875)^2+ (62-65.875)^2+ (65-65.875)^2+( 61-65.875)^2+ (70-65.875)^2+ (66-65.875)^2+ (66-65.875)^2+ (71+63-65.875)^2+ (69-65.875)^2+ (65-65.875)^2+ (71-65.875)^2+( 66-65.875)^2+ (66-65.875)^2+ (69-65.875)^2+ (68-65.875)^2+ (62-65.875)^2+ (65-65.875)^2+ (67-65.875)^2,+\\(65-65.875)^2+ (71-65.875)^2+ (65-65.875)^2+ (70-65.875)^2+( 62-65.875)^2+( 62-65.875)^2+ (63-65.875)^2+ (64-65.875)^2+ (67-65.875)^2+ (67-65.875)^2 / 32[/tex]

=>[tex]v= 8.567329[/tex]

=>   [tex]\sigma = \sqrt{8.567329}[/tex]

=>   [tex]\sigma = 2.927[/tex]

The level of significance is evaluated as

        [tex]\alpha = 1 - 0.95[/tex]

        [tex]\alpha = 0.05[/tex]

The degree of freedom is  evaluated as

    [tex]Df = n- 1 \equiv Df = 32 -1 = 31[/tex]

The critical values for the level of significance is obtained from the z -table as

      [tex]t_c = t_{\alpha/2 } , Df = t _{0.05/2}, 31 =\pm 1.96[/tex]

The confidence interval is evaluated as

       [tex]\mu = \= x \pm t_c * \frac{\sigma }{\sqrt{n} }[/tex]

substituting values

        [tex]\mu =65.875 \pm 1.96* \frac{2.927}{\sqrt{32} }[/tex]

       [tex]\mu =65.875 \pm 1.01415[/tex]

=>    [tex]64.86<\mu<67[/tex]

"The counting rule that is used for counting the number of experimental outcomes when n objects are selected from a set of N objects where order of selection is not important is called the"

Answers

Options

Counting rule for permutationsCounting rule for multiple-step experimentsCounting rule for combinationsCounting rule for independent events

Answer:

(C)Counting rule for combinations

Step-by-step explanation:

When selecting n objects from a set of N objects, we can determine the number of experimental outcomes using permutation or combination.

When the order of selection is important, we use permutation.However, whenever the order of selection is not important, we use combination.

Therefore, The counting rule that is used for counting the number of experimental outcomes when n objects are selected from a set of N objects where order of selection is not important is called the counting rule for combinations.

A selection or listing of objects in which the order of the objects is not important

Factor the expression. 9b2 – 25 a. (5b + 3)(5b – 3) b. (3b – 5)(3b – 5) c. (3b + 5)(3b + 5) d. (3b + 5)(3b – 5)

Answers

Answer:

Dear user,

Answer to your query is provided below

Option (d) is correct.

Step-by-step explanation:

Explanation of the same is attached in image

Use the fraction strips to compare 2/4 and 5/8. use the drop-down menus to explain your comparison.


Comparing 2/4 and 5/8 is the same as comparing the length of ____ of the 1/4 sections and _____ of the 1/8 sections.

The sections of 2/4 is _____ the length of 5/8. Therefore, 2/4 < 5/8.

Answers

Answer:

2

5

[tex]\dfrac{4}{5}[/tex]

Step-by-step explanation:

We are given 2 fractional numbers:

[tex]1.\ \dfrac{2}4\\2.\ \dfrac{5}8[/tex]

We have to use fraction strips to compare to the fractional numbers.

Let we are Comparing [tex]\frac{2}{4}[/tex] with the length of [tex]x[/tex] number of [tex]\frac{1}4[/tex] sections.

i.e.

[tex]\dfrac{2}{4} = x \times \dfrac{1}{4}\\\Rightarrow x = \dfrac{2 \times 4}{4}\\\Rightarrow x = 2[/tex]

Let we are Comparing [tex]\frac{5}{8}[/tex] with the length of [tex]y[/tex] number of [tex]\frac{1}8[/tex] sections.

i.e.

[tex]\dfrac{5}{8} = y \times \dfrac{1}{8}\\\Rightarrow y = \dfrac{5 \times 8}{8}\\\Rightarrow y = 5[/tex]

Now, let us have a look at 3rd part of question:

The sections of 2/4 is _____ the length of 5/8. Therefore, 2/4 < 5/8

Let the answer be [tex]z[/tex].

So, the equation becomes:

[tex]\dfrac{2}{4} = z \times \dfrac{5}{8}\\\Rightarrow z = \dfrac{2 \times 8}{4 \times 5}\\\Rightarrow z = \dfrac{2 \times 2}{5}\\\Rightarrow z = \dfrac{4}{5}[/tex]

So, the answers are:

2

5

[tex]\dfrac{4}{5}[/tex]

I need help on khan academy and I’m willing to pay half right when you start the work and when you finish all the work I’ll pay the other half. I am desperate if I don’t pass this class it’s bad for me. Comment your number,snap anything. If not can you please help me with this one problem please thank youu

Answers

Answer:

[tex]x\approx 50^\circ[/tex]

Step-by-step explanation:

[tex]c^2 = a^2 + b^2 - 2ab(cos(C))[/tex]

See the figure below to get the values as:

[tex]7^2=7^2+9^2-2\left(7\right)\left(9\right)cos\left(x\right)\\\\cos(x)=\frac{7^2+9^2-7^2}{2\cdot \:7\cdot \:9}\\\\x\approx 50^\circ[/tex]

There are multiple concepts to solve this problem. This is one of the concept used in high school. Other concept to solve this problem is to use the concept of isosceles triangle. An isosceles triangle is a triangle with (at least) two equal sides. The angles shared by the two equal sides are also equal. So that the sum of all the three angles will add up to 180.

[tex]x+x+80=180\\\\2x=100\\\\x=50^{\circ}[/tex]

Best Regards!

In converting 750 ounces to pounds, what unit (omit the number) would you
place in the denominator of your ratio? Use the plural form in your answer.
Remember that there are 16 ounces in 1 pound.
Answer here
SUBMIT

Answers

16 ounces is 1 pound.

So 1 ounce will be 1/16 pound.

750 × 1/16

[tex]\displaystyle \frac{750}{16}[/tex]

Answer:

The correct answer is ounces

Step-by-step explanation:

1 pound= 16 ounces

750x 1/16=7.50

so it will be ounces

Hope this helps!

The histogram shows the number of miles driven by a sample of automobiles in New York City.


What is the minimum possible number of miles traveled by an automobile included in the histogram?

Answers

Answer:

0 miles

Step-by-step explanation:

The computation of the minimum possible number of miles traveled by  automobile is shown below:

As we can see that in the given histogram it does not represent any normal value i.e it is not evenly distributed moreover, the normal distribution is symmetric that contains evenly distribution data

But this histogram shows the asymmetric normal distribution that does not have evenly distribution data

Therefore the correct answer is 0 miles

Answer:

2,500

That is your correct answer.

Please show you work for both 5 and 6!

Answers

Answer:

5-c)  6-c)

Step-by-step explanation:

5 c)= (a^4)^(1/3)= a^(4*1/3)=a^(4/3)

6. At the beginning of 1st year the total value on the bank account=500 USD

At the end of 1-st year the total value is 500 USD +3% from 500 USD=

500+500*0.03= 500*1.03

At the end of 2-nd year the total amount is 500*1.03+3% from 500*1.03=

500*1.03^2

Similarly at the end of the 3-rd year the total amount is 500*1.03^3

Finally at the end of fourth year the total amount is 500*1.03^4

Please answer this correctly

Answers

Answer:

6 pizzas

Step-by-step explanation:

At least 10 and fewer than 20 makes it 10-19

So,

10-19 => 6 pizzas

6 pizzas have at least 10 pieces of pepperoni but fewer than 20 pieces of pepperoni.

A tank contains 70 lb of salt dissolved in 400 gallons of water. A brine solution is pumped into the tank at a rate of 4 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 4 gal/min. Determine A(t), the amount of salt in the tank at time t, if the concentration of salt in the inflow is variable and given by cin(t)

Answers

Answer:

[tex]A(t)=400C_{in}(t)+[70-400C_{in}(t)]\cdot e^{-\frac{t}{100}}[/tex]

Step-by-step explanation:

Volume of water in the Tank =400 gallons

Let A(t) be the amount of salt in the tank at time t.

Initially, the tank contains 70 lbs of salt, therefore:

A(0)=70 lbs

Amount of Salt in the Tank

[tex]\dfrac{dA}{dt}=R_{in}-R_{out}[/tex]

=(concentration of salt in inflow)(input rate of brine)

[tex]=(C_{in}(t))( 4\frac{gal}{min})\\=4C_{in}(t)\frac{lbs}{min}[/tex]

=(concentration of salt in outflow)(output rate of brine)

[tex]=(\frac{A(t)}{400})( 4\frac{gal}{min})=\frac{A}{100}[/tex]

Therefore:

[tex]\dfrac{dA}{dt}=4C_{in}(t)-\dfrac{A}{100}[/tex]

We then solve the resulting differential equation by separation of variables.

[tex]\dfrac{dA}{dt}+\dfrac{A}{100}=4C_{in}(t)\\$The integrating factor: e^{\int \frac{1}{100}}dt =e^{\frac{t}{100}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{100}}+\dfrac{A}{100}e^{\frac{t}{100}}=4C_{in}(t)e^{\frac{t}{100}}\\(Ae^{\frac{t}{100}})'=4C_{in}(t)e^{\frac{t}{100}}[/tex]

Taking the integral of both sides

[tex]\int(Ae^{\frac{t}{100}})'=\int [4C_{in}(t)e^{\frac{t}{100}}]dt\\Ae^{\frac{t}{100}}=4*100C_{in}(t)e^{\frac{t}{100}}+C, $(C a constant of integration)\\Ae^{\frac{t}{100}}=400C_{in}(t)e^{\frac{t}{100}}+C\\$Divide all through by e^{\frac{t}{100}}\\A(t)=400C_{in}(t)+Ce^{-\frac{t}{100}}[/tex]

Recall that when t=0, A(t)=70 lbs (our initial condition)

[tex]A(t)=400C_{in}(t)+Ce^{-\frac{t}{100}}\\

70=400C_{in}(t)+Ce^{-\frac{0}{100}}\\

C=70-400C_{in}(t)\\$Therefore, the amount of salt in the tank at any time t is:

\\\\A(t)=400C_{in}(t)+[70-400C_{in}(t)]\cdot e^{-\frac{t}{100}}[/tex]

triangle STU is similar to triangle VWX. Which statement is true about the two triangles?​

Answers

Answer:

Option 1.

Step-by-step explanation:

When triangles are similar, their angles cannot be proportional. The angles on both triangles have to be same.

Option 3 and 4 are wrong.

Angle T and angle U cannot be congruent on the same triangle.

Therefore, option 1 is correct.

The answer would be the third one because if they are simillar  that means they are not exactly the same but one is a dillation of one. This means they are proportinate. Mark Branliest!!!!

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