On average, the printer uses 500 sheets of paper each day with a standard deviation of 10 sheets. What is the probability that the printer uses more than 508 sheets?

Answer:

216 cm³

Step-by-step explanation:

The volume of a cube is denoted by V = s³, where s is the side length.

Here, the side length is 6 centimetres, so plug this into the formula to find V:

V = s³

V = 6³ = 6 * 6 * 6 = 216

The answer is thus 216 cm³.

~ an aesthetics lover

Answer:

216

Step-by-step explanation:

6³ = 216

Answer:

2{3[9+4(7-5)-4]}

2{3[9+4(2)-4]}

2{3[13(2)-4]}

2{3[26-4]}

2{3[22]}

2{66}

132

Step-by-step explanation:

Answer:

a) 32

b) none?

c) C & F

D) 9/5, 32?

Step-by-step explanation:

Answer:

6 mg

Step-by-step explanation:

12 mg

0.018 g * 1000 mg/g = 18 mg

18 mg - 12 mg = 6 mg

We assume that question b is asking for the distribution of [tex] \\ \overline{x}[/tex], that is, the distribution for the average amount of pollutants.

Answer:

a. The distribution of X is a normal distribution [tex] \\ X \sim N(8.6, 1.3)[/tex].

b. The distribution for the average amount of pollutants is [tex] \\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}})[/tex].

c. [tex] \\ P(z>-0.08) = 0.5319[/tex].

d. [tex] \\ P(z>-0.47) = 0.6808[/tex].

e. We do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for [tex] \\ \overline{X}[/tex] is also normal because the sample was taken from a normal distribution.

f. [tex] \\ IQR = 0.2868[/tex] ppm. [tex] \\ Q1 = 8.4566[/tex] ppm and [tex] \\ Q3 = 8.7434[/tex] ppm.

Step-by-step explanation:

First, we have all this information from the question:

a. What is the distribution of X?

The distribution of X is the normal (or Gaussian) distribution. X (uppercase) is the random variable, and follows a normal distribution with [tex] \\ \mu = 8.6[/tex] ppm and [tex] \\ \sigma =1.3[/tex] ppm or [tex] \\ X \sim N(8.6, 1.3)[/tex].

b. What is the distribution of [tex] \\ \overline{x}[/tex]?

The distribution for [tex] \\ \overline{x}[/tex] is [tex] \\ N(\mu, \frac{\sigma}{\sqrt{n}})[/tex], i.e., the distribution for the sampling distribution of the means follows a normal distribution:

[tex] \\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}})[/tex].

c. What is the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants?

Notice that the question is asking for the random variable X (and not [tex] \\ \overline{x}[/tex]). Then, we can use a standardized value or z-score so that we can consult the standard normal table.

[tex] \\ z = \frac{x - \mu}{\sigma}[/tex] [1]

x = 8.5 ppm and the question is about [tex] \\ P(x>8.5)[/tex]=?  

Using [1]

[tex] \\ z = \frac{8.5 - 8.6}{1.3}[/tex]

[tex] \\ z = \frac{-0.1}{1.3}[/tex]

[tex] \\ z = -0.07692 \approx -0.08[/tex] (standard normal table has entries for two decimals places for z).

For [tex] \\ z = -0.08[/tex], is [tex] \\ P(z<-0.08) = 0.46812 \approx 0.4681[/tex].

But, we are asked for [tex] \\ P(z>-0.08) \approx P(x>8.5)[/tex].

[tex] \\ P(z<-0.08) + P(z>-0.08) = 1[/tex]

[tex] \\ P(z>-0.08) = 1 - P(z<-0.08)[/tex]

[tex] \\ P(z>-0.08) = 0.5319[/tex]

Thus, "the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants" is [tex] \\ P(z>-0.08) = 0.5319[/tex].

d. For the 38 cities, find the probability that the average amount of pollutants is more than 8.5 ppm.

Or [tex] \\ P(\overline{x} > 8.5)[/tex]ppm?

This random variable follows a standardized random variable normally distributed, i.e. [tex] \\ Z \sim N(0, 1)[/tex]:

[tex] \\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex] [2]

[tex] \\ z = \frac{\overline{8.5} - 8.6}{\frac{1.3}{\sqrt{38}}}[/tex]

[tex] \\ z = \frac{-0.1}{0.21088}[/tex]

[tex] \\ z = \frac{-0.1}{0.21088} \approx -0.47420 \approx -0.47[/tex]

[tex] \\ P(z<-0.47) = 0.31918 \approx 0.3192[/tex]

Again, we are asked for [tex] \\ P(z>-0.47)[/tex], then

[tex] \\ P(z>-0.47) = 1 - P(z<-0.47)[/tex]

[tex] \\ P(z>-0.47) = 1 - 0.3192[/tex]

[tex] \\ P(z>-0.47) = 0.6808[/tex]

Then, the probability that the average amount of pollutants is more than 8.5 ppm for the 38 cities is [tex] \\ P(z>-0.47) = 0.6808[/tex].

e. For part d), is the assumption that the distribution is normal necessary?

For this question, we do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for [tex] \\ \overline{X}[/tex] is also normal because the sample was taken from a normal distribution. Additionally, the sample size is large enough to show a bell-shaped distribution.  

f. Find the IQR for the average of 38 cities.

We must find the first quartile (25th percentile), and the third quartile (75th percentile). For [tex]\\ P(z<0.25)[/tex], [tex] \\ z \approx -0.68[/tex], then, using [2]:

[tex] \\ -0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}[/tex]

[tex] \\ (-0.68 *0.21088) + 8.6 = \overline{X}[/tex]

[tex] \\ \overline{x} =8.4566[/tex]

[tex] \\ Q1 = 8.4566[/tex] ppm.

For Q3

[tex] \\ 0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}[/tex]

[tex] \\ (0.68 *0.21088) + 8.6 = \overline{X}[/tex]

[tex] \\ \overline{x} =8.7434[/tex]

[tex] \\ Q3 = 8.7434[/tex] ppm.

[tex] \\ IQR = Q3-Q1 = 8.7434 - 8.4566 = 0.2868[/tex] ppm

Therefore, the IQR for the average of 38 cities is [tex] \\ IQR = 0.2868[/tex] ppm. [tex] \\ Q1 = 8.4566[/tex] ppm and [tex] \\ Q3 = 8.7434[/tex] ppm.

Answer:

The answer is option A.

Step-by-step explanation:

2 - 35 can be written as 2 + (-35) since in 2+( -35) when the bracket is removed it becomes 2 - 35.

Hope this helps

Answer:

Your correct answer is option a. 2 + (-35)

Step-by-step explanation:

When you are finding which equation is equivalent to the one you have, the best thing to find is what your equation's answer is.

Answer:

1/90 = 1.11%

Step-by-step explanation:

We have that the number of ways of total selections and assignments possible is a permutation.

We know that permutations are defined like this:

nPr = n! / (n-r)!

In our case n = 10 and r = 2, replacing:

10P2 = 10! / (10 - 2)! = 10! / 8!

10P2 = 90

In addition to this, there will only be one way to randomly select the two members currently holding the positions of President and Vice President and reassign them to their current positions. Thus,

Probability would come being the following:

P = 1/90 = 1.11%

Answer: B. Sameer did not use the correct equation

Step-by-step explanation:

12 IS three-fourths OF x

IS: equals

OF: multiplication

[tex]12=\dfrac{3}{4}x[/tex]

48 = 3x

16 = x

Answer:

it's b in Edg

Step-by-step explanation:

The point that is a solution to the inequality shown in the graph is:

A. (0,5).

The points that are on the region shaded in blue are solutions to the inequality.

(3,2) and (-3,-6) are on the dashed line, hence they are not solutions. Point (5,0) is to the right of the line, hence it is not a solution, and point (0,5) is a solution, meaning that option A is correct.

More can be learned about inequalities at https://brainly.com/question/25235995

#SPJ1

Step-by-step explanation:

input x , output y

if x= x1 then y=y1 and y1 is the only value then it is a function

if we get multiple values of y then it is not a function

Answer:

The critical value of  [tex]f(\theta) = 6\cdot \cos \theta + 3\cdot \sin 2\theta[/tex] are given by [tex]\theta \approx 0.091\pi \pm 2\pi\cdot n[/tex] or [tex]\theta \approx 0.909\pi \pm 2\pi \cdot n[/tex], [tex]\forall \,n \in \mathbb{N}[/tex]

Step-by-step explanation:

The function to be evaluated is [tex]f(\theta) = 6\cdot \cos \theta + 3\cdot \sin 2\theta[/tex], the first derivative of the function must be taken in order to determine the set of critical numbers. Each derivative are found by using the differentiation rule for a sum of functions and rule of chain and subsequently simplified by trigonometric and algebraic means:

First derivative

[tex]f'(\theta) = - 6 \cdot \sin \theta +6\cdot \cos 2\theta[/tex]

[tex]f'(\theta) = -6\cdot \sin \theta + 2\cdot (\cos^{2}\theta-\sin^{2}\theta)[/tex]

[tex]f'(\theta) = -6\cdot \sin \theta + 2\cdot [(1-\sin^{2}\theta-\sin^{2}\theta)][/tex]

[tex]f'(\theta) = -6\cdot \sin \theta + 2\cdot (1-2\cdot \sin^{2}\theta)[/tex]

[tex]f'(\theta) = -6\cdot \sin \theta + 2 - 4\cdot \sin^{2}\theta[/tex]

[tex]f'(\theta) = -4\cdot \sin^{2}\theta - 6\cdot \sin \theta +2[/tex]

The procedure to determine the critical number of the given function are described briefly:

1) First derivative is equalised to zero.

2) The resultant equation is solved.

Then,

[tex]-4\cdot \sin^{2}\theta - 6\cdot \sin \theta +2 = 0[/tex]

Whose roots are:

[tex]\sin \theta_{1} \approx 0.281[/tex] and [tex]\sin \theta_{2} \approx -1.781[/tex]

The sine function is a continuous function with a range between 1 and -1, so, only the first root offers a realistic solution. In addition, such function is positive at first and second quadrants and has a periodicity of [tex]2\pi[/tex] radians, the family of critical values are determined by the unse of inverse trigonometric functions:

[tex]\theta \approx \sin^{-1} 0.281[/tex]

There are two subsets of solutions:

[tex]\theta \approx 0.091\pi \pm 2\pi\cdot n[/tex] or [tex]\theta \approx 0.909\pi \pm 2\pi \cdot n[/tex], [tex]\forall \,n \in \mathbb{N}[/tex]

Answer:

b. 1.15

Step-by-step explanation:

The z statistics is given by:

[tex]Z = \frac{X - p}{s}[/tex]

In which X is the found proportion, p is the expected proportion, and s, which is the standard error is [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

Out of 53 teachers who replied to the survey, 13 claim they are satisfied with their job.

This means that [tex]X = \frac{13}{53} = 0.2453[/tex]

She find that the proportion of satisfied teachers nationally is 18.4%.

This means that [tex]p = 0.184[/tex]

Standard error:

p = 0.184, n = 53.

So

[tex]s = \sqrt{\frac{0.184*0.816}{53}} = 0.0532[/tex]

Z-statistic:

[tex]Z = \frac{X - p}{s}[/tex]

[tex]Z = \frac{0.2453 - 0.184}{0.0532}[/tex]

[tex]Z = 1.15[/tex]

The correct answer is:

b. 1.15

Answer:

when a straight line cuts two or more parallel lines then the angles forming on the side of transversal line exteriorly opposite to eachother is called exterior alternative angle.

for eg if AB //CD and EF is a transversal line meeting the parallel lines at G abd H then the exterior alternative angle are angle EGB = angle CHF and angle AGE=angle DHF are two pairs of exterior alternative angle .

hope its helpful to uh !!!!!!

Answer:

9

Step-by-step explanation:

p^2 -4p +4

Let p = -1

(-1)^1 -4(-1) +4

1 +4+4

9

Answer:

[tex]y=31x^2+77x+41[/tex]

which agrees with option C in your list of possible answers.

Step-by-step explanation:

Since normally functions are represented on the x-y plane, it is common to replace h(x) with the "y" variable of the vertical axis where its values will be represented (plotted). Then the expression can be also written as follows:

[tex]h(x)=31x^2+77x+41\\y=31x^2+77x+41[/tex]

Your answer would be 76.2% to the nearest tenth.

We can find this by first dividing 16 by 21 to get 0.7619. which is the proportion as a decimal. To convert this into a percentage, we need to multiply it by 100 to get 76.19% = 76.2% to the nearest tenth.

I hope this helps! Let me know if you have any questions :)

Answer:

a) [tex]P(X=2)=(4C2)(0.375)^2 (1-0.375)^{4-2}=0.330[/tex]  

b) [tex]P(X\geq 2)=1-P(X< 2)=1-[P(X=0)+P(X=1)][/tex]

[tex]P(X=0)=(4C0)(0.375)^0 (1-0.375)^{4-0}=0.153[/tex]  

[tex]P(X=1)=(4C1)(0.375)^1 (1-0.375)^{4-1}=0.366[/tex]  

And replacing we got:

[tex]P(X\geq 2)=1-P(X< 2)=1-[0.153+0.366]=0.481[/tex]

Step-by-step explanation:

Let X the random variable of interest, on this case we now that:  

[tex]X \sim Binom(n=4, p=0.375)[/tex]  

The probability mass function for the Binomial distribution is given as:  

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]  

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

Part a

[tex]P(X=2)=(4C2)(0.375)^2 (1-0.375)^{4-2}=0.330[/tex]  

Part b

[tex]P(X\geq 2)=1-P(X< 2)=1-[P(X=0)+P(X=1)][/tex]

[tex]P(X=0)=(4C0)(0.375)^0 (1-0.375)^{4-0}=0.153[/tex]  

[tex]P(X=1)=(4C1)(0.375)^1 (1-0.375)^{4-1}=0.366[/tex]  

And replacing we got:

[tex]P(X\geq 2)=1-P(X< 2)=1-[0.153+0.366]=0.481[/tex]

Answer:

$2.06

Step-by-step explanation:

$2.99 x 6 = $17.94

$20.00 - $17.94 = $2.06

Hope this helps

Answer: $0.26

Step-by-step explanation:

Cost of 6 pens

= 2.99 x 6

= 17.94

Add sales tax at 10%,

= 17.94 x 1.1

= 19.74

Change due to me

= 20 - 19.74

= 0.26

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable X represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), p = 0.52

The random variable X thus follows a Binomial distribution with parameters n = 5 and p = 0.52.

(1)

Compute the probability of overbooking  as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

                                        [tex]=P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135[/tex]

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

[tex]=P(X<3)\\\\1-P(X\geq 3)\\\\=1-[{5\choose 3}(0.52)^{3}(1-0.52)^{5-3}+{5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}]\\\\=1-[0.323960832+0.175478784+0.0380204032]\\\\=0.4625399808\\\\\approx 0.4625[/tex]

Thus, the probability that the flight has empty seats is 0.4625.

Answer:

This question is very simple,

Ok first you will need to find the x and y intercepts by letting y=0 and x=0

First let x=0

so, 0-y=1

y=-1

let y=0

x-0=1

x=1

now we know

x-intercept=(1,0)

y-intercept=(0-1)

Hence, find the graph that has the two corresponding points and that would be the graph you are looking for.

Step-by-step explanation:

Answer:

1144cm²

Approx. 1100cm²

Step-by-step explanation:

Area of rectangle=

L×B

34×20 = 680cm²

Area of semi-circle=

πr²

3.14 × (10)² = 314cm²

Area of triangle=

½b×h

b = 49 - 34 = 15

20 × 15 = 150cm²

2

Area of shape =

680 + 314 + 150 = 1144cm²

approx. 1100cm²

Please mark my answer as brainliest <3

Answer:

The solution:

X = 16 and Y = -6

Step-by-step explanation:

The equations to be solved are:

x+y = 10 ------- equation 1

x+2y = 4 ----------- equation 2

we can multiply equation 1 by -1 to make the value of x and y negative.

This will give us

-x- y = - 10 ------- equation 3

x+2y = 4 ----------- equation 2

We will now add equations 3 and 2 together so that x will cancel itself out.

this will give us

y = -10 +4 = -6

hence, we have the value of y as -6.

To get the value of x, we can put this value of y into any of the equations above.  (I will use equation 1)

x - 6 = 10

from this, we have that x = 4

Therefore, we have our answer as

X = 16 and Y = -6

Answer:

y= 3/13x + 2/13

Step-by-step explanation:

3x-13y=2

Subtract 3x from both side

-13y=-3x-2

Divide by -13

y= 3/13x + 2/13

Answer:

[tex]y = \frac{2-3x}{-13}[/tex]

Step-by-step explanation:

=> 3x-13y = 2

Subtract 3x to both sides

=> -13y = 2-3x

Dividing both sides by -13

=> [tex]y = \frac{2-3x}{-13}[/tex]

Answer:

  82.5%

Step-by-step explanation:

It helps to start with the correct formula:

  f(x) = ((6)(1.0) +x(0.3))/(6 +x) . . . . parentheses are required

Then f(2) is ...

  f(2) = (6 +.3(2))/(6+2) = 6.6/8

  f(2) = 82.5%

Answer:

7 to 15, 7:15, 7/15

Step-by-step explanation:

Ratios can be written as:

a to b

a:b

a/b

We want to find the ratio of exercise days to days in the month. She exercises 14 days out of 30 days in the month. Therefore,

a= 14

b= 30

14 to 30

14:30

14/30

The ratios can be simplified. Both numbers can be evenly divided by 2.

(14/2) to (30/2)

7 to 15

(14/2) : (30/2)

7:15

(14/2) / (30/2)

7/15

Answer:

divide both numbers by 14.. the ans is 1: 2

Answers:

a) 0.0625

b) 0.9375

==================================================

Work Shown:

The probability of landing on heads is 1/2 = 0.5 since both sides are equally likely to land on. Getting 4 heads in a row is (1/2)^4 = (0.5)^4 = 0.0625

The event of getting at least one tail is the complement of getting all four heads. This is because you either get all four heads or you get at least one tail. One or the other must happen. We subtract the result we got from 1 to get 1-0.0625 = 0.9375

You can think of it like this

P(getting all four heads) + P(getting at least one tail) = 1

The phrasing "at least one tail" means "one tail or more".

Answer:

The answer to the best prediction is 115.04

Step-by-step explanation:

We have to:

x = 102

They also tell us that:

y = 5.9 + 1.07 * x

If we replace we have:

y = 5.9 + 1.07 * (102)

y = 115.04

Therefore, the best predicted value of ModifyingAbove and with caret given that the older child has an IQ of 102 is 115.04

Answer:

a) Null and alternative hypothesis

[tex]H_0: \mu=32.48\\\\H_a:\mu< 32.48[/tex]

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

[tex]H_0: \mu=32.48\\\\H_a:\mu< 32.48[/tex]

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

[tex]H_0: \mu=32.48\\\\H_a:\mu< 32.48[/tex]

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

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

[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884[/tex]

Then, we can calculate the t-statistic as:

[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=65-1=64[/tex]

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

[tex]\text{P-value}=P(t<-1.565)=0.0612[/tex]

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Critical value approach

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.