Point c (2,2) is the center of the circle. what is the ratio of ac to the length of dc?

1) 1:2
2) 2:1
3) 1:1
4) 3:1

Answer:

f(x) = -3x + 4

Step-by-step explanation:

Step 1: Move the 9x over

3y = 12 - 9x

Step 2: Divide everything by 3

y = 4 - 3x

Step 3: Rearrange

y = -3x + 4

Step 4: Change y to f(x)

f(x) = -3x + 4

Answer:

The answer is "Option A"

Step-by-step explanation:

The valid linear programming language equation can be defined as follows:

Equation:

[tex]\Rightarrow \ Min\ 4x + 3y + (\frac{2}{3})z[/tex]

The description of a linear equation can be defined as follows:

It is an algebraic expression whereby each term contains a single exponent, and a single direction consists in the linear interpolation of the equation.

Formula:

[tex]\to \boxed{y= mx+c}[/tex]

Answer:

The frequency table is shown below.

Step-by-step explanation:

The data set arranged ascending order is:

S = {33 , 34 , 39 , 48 , 49 , 50 , 53 , 54 , 55 , 56 , 58 , 58,  60 , 63 , 64 , 65 , 70 , 71 , 74 , 84}

It is asked to use the minimum value from the data set as the lower class limit for the first row.

So, the lower class limit for the first class interval is 33.

To determine the class width compute the range as follows:

[tex]\text{Range}=\text{Maximum}-\text{Minimum}[/tex]

          [tex]=84-33\\=51[/tex]

The number of classes requires is 5.

The class width is:

[tex]\text{Class width}=\frac{Range}{5}=\frac{51}{2}=10.2\approx 10[/tex]

So, the class width is 10.

The classes are:

33 - 42

43 - 52

53 - 62

63 - 72

73 - 82

83 - 92

Compute the frequencies of each class as follows:

Class Interval                  Values                        Frequency

   33 - 42                      33 , 34 , 39                             3

   43 - 52                      48 , 49 , 50                            3

   53 - 62          53 , 54 , 55 , 56 , 58 , 58,  60              7

   63 - 72                 63 , 64 , 65 , 70 , 71                      5

   73 - 82                              74                                  1

   83 - 92                             84                                   1

   TOTAL                                                                   20

Compute the relative frequencies as follows:

Class Interval          Frequency        Relative Frequency

   33 - 42                        3                   [tex]\frac{3}{20}\times 100\%=15\%[/tex]

   43 - 52                        3                   [tex]\frac{3}{20}\times 100\%=15\%[/tex]

   53 - 62                        7                   [tex]\frac{7}{20}\times 100\%=35\%[/tex]

   63 - 72                        5                   [tex]\frac{5}{20}\times 100\%=25\%[/tex]

   73 - 82                         1                   [tex]\frac{1}{20}\times 100\%=5\%[/tex]

   83 - 92                         1                   [tex]\frac{1}{20}\times 100\%=5\%[/tex]

   TOTAL                        20                          100%

Answer:

Mia:10 Maya:5 Maria:3

Step-by-step explanation:

3+7= 10= Mia's age

10÷2=5= Maya's age

Answer:

Mia - 10

Maya - 5

Maria - 3

Answer:

The test to be used is the right tailed test.

Step-by-step explanation:

The type of test joe should do would be a right tailed test. This is because;

A right tailed test which we sometimes call an upper test is where the hypothesis statement contains the greater than (>) symbol. This means that, the inequality points to the right. For example, we want to compare the the life of batteries before and after a manufacturing change.

If we want to know if the battery life of maybe 90 hours would be greater than the original, then our hypothesis statements might be:

Null hypothesis: (H0 = 90).

Alternative hypothesis: (H1) > 90.

In the null hypothesis, there are no changes, but in the alternative hypothesis, the battery life in hours has increased.

So, the most important factor here is that the alternative hypothesis (H1) is what determines if we have a right tailed test, not the null hypothesis.

Thus, the test to be used is the right tailed test.

Answer:

right tailed test.

Step-by-step explanation:

Answer:

7a+2b-3c

Step-by-step explanation:

6a+a = 7a

2b stays the same

-3c stays the same

Answer:

Hey mate, here is your answer. Hope it helps you.

7a-3c+2b

Step-by-step explanation:

6a+a-3c+2b

=7a-3c+2b

3c and 2b will be the same because the variables are different. They are not like terms.

Answer:

One solution

Step-by-step explanation:

99% of the time, linear equations (equations that have the first degree) have only one solution. However, it's always good to check.

6 - 3x = 12 - 6x

6 = 12 - 3x

-3x = -6

x = 2

As you can see, only one solution. Hope this helps!

Answer:

55.11% probability that the mean life a random sample of 9 such blenders falls between 4.5 and 5.1 years.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

[tex]\mu = 5, \sigma = 1, n = 9, s = \frac{1}{\sqrt{9}} = 0.3333[/tex]

Find the probability that the mean life a random sample of 9 such blenders falls between 4.5 and 5.1 years.

This is the pvalue of Z when X = 5.1 subtracted by the pvalue of Z when X = 4.5. So

X = 5.1

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

By the Central Limit Theorem

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

[tex]Z = \frac{5.1 - 5}{0.3333}[/tex]

[tex]Z = 0.3[/tex]

[tex]Z = 0.3[/tex] has a pvalue of 0.6179

X = 4.5

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

[tex]Z = \frac{4.5 - 5}{0.3333}[/tex]

[tex]Z = -1.5[/tex]

[tex]Z = -1.5[/tex] has a pvalue of 0.0668

0.6179 - 0.0668 = 0.5511

55.11% probability that the mean life a random sample of 9 such blenders falls between 4.5 and 5.1 years.

Circumference = πd

~substitute → (π)(6 cm)

~simplify → 6π cm.

So the circumference of the circle shown here is 6π cm.

Answer:

18.85 cm

Step-by-step explanation:

The circumference of a circle has a formula.

Circumference = π × diameter

The diameter is 6 centimeters.

Circumference = π × 6

Circumference ≈ 18.85

The circumference of the circle is 18.85 centimeters.

Answer:

Following are the description of the given course code:

Step-by-step explanation:

The given course code is Pre-Algebra, which is just an introduction arithmetic course programs to train high school in the Algebra 1. This course aims to strengthen required problem solving skills, datatypes, equations, as well as graphing.

Answer:

A

Step-by-step explanation:

Answer:

a) Mean = 0.125 inch

Standard deviation = 0.13975 inch

b) Probability that the width of the casing minus the width of the door exceeds 1/4 inch = P(X > 0.25) = 0.18673

c) Probability that the door does not fit in the casing = P(X < 0) = 0.18673

Step-by-step explanation:

Let the distribution of the width of the casing be X₁ (μ₁, σ₁²)

Let the distribution of the width of the door be X₂ (μ₂, σ₂²)

The distribution of the difference between the width of the casing and the width of the door = X = X₁ - X₂

when two independent normal distributions are combined in any manner, the resulting distribution is also a normal distribution with

Mean = Σλᵢμᵢ

λᵢ = coefficient of each disteibution in the manner that they are combined

μᵢ = Mean of each distribution

Combined variance = σ² = Σλᵢ²σᵢ²

λ₁ = 1, λ₂ = -1

μ₁ = 24 inches

μ₂ = 23 7/8 inches = 23.875 inches

σ₁² = (1/8)² = (1/64) = 0.015625

σ₂ ² = (1/16)² = (1/256) = 0.00390625

Combined mean = μ = 24 - 23.875 = 0.125 inch

Combined variance = σ² = (1² × 0.015625) + [(-1)² × 0.00390625] = 0.01953125

Standard deviation = √(Variance) = √(0.01953125) = 0.1397542486 = 0.13975 inch

b) Probability that the width of the casing minus the width of the door exceeds 1/4 inch = P(X > 0.25)

This is a normal distribution problem

Mean = μ = 0.125 inch

Standard deviation = σ = 0.13975 inch

We first normalize/standardize 0.25 inch

The standardized score of any value is that value minus the mean divided by the standard deviation.

z = (x - μ)/σ = (0.25 - 0.125)/0.13975 = 0.89

P(X > 0.25) = P(z > 0.89)

Checking the tables

P(x > 0.25) = P(z > 0.89) = 1 - P(z ≤ 0.89) = 1 - 0.81327 = 0.18673

c) Probability that the door does not fit in the casing

If X₂ > X₁, X < 0

P(X < 0)

We first normalize/standardize 0 inch

z = (x - μ)/σ = (0 - 0.125)/0.13975 = -0.89

P(X < 0) = P(z < -0.89)

Checking the tables

P(X < 0) = P(z < -0.89) = 0.18673

Hope this Helps!!!

Answer:

right triangle

Step-by-step explanation:

We can use the Pythagorean theorem to determine if this is a right triangle

a^2 + b^2 = c^2

13^2 + ( 8 sqrt(13)) ^2 = (sqrt(1001))^2

169 + 8^2 * 13 = 1001

169+64*13 = 1001

169+832=1001

1001 = 1001

Since this is true, this is a right triangle

Answer:

The degrees of freedom are given by:

[tex] df =n-1= 15-1=14[/tex]

And if we look in the chi square distribution with 14 degrees of freedom and if we find a quantile who accumulates 0.1 of the area in the left we got:

[tex] \chi^2 = 7.790[/tex]

And then the best answer would be:

c. 7.790

Step-by-step explanation:

For this case we know that we are using a one tailed (lower tail) critical value using a significance level of [tex]\alpha=0.1[/tex] and for this case we know that the ample size is n=15. The degrees of freedom are given by:

[tex] df =n-1= 15-1=14[/tex]

And if we look in the chi square distribution with 14 degrees of freedom and if we find a quantile who accumulates 0.1 of the area in the left we got:

[tex] \chi^2 = 7.790[/tex]

And then the best answer would be:

c. 7.790

Answer:

see explanations

Step-by-step explanation:

The given blue line has a slope of m = -1/2.

The line parallel to the given line passing through point (x0,y0)=(2,3) is given by the point-slope form:

(y-y0)=m(x-x0)

substitute values

(y-3) = (-1/2)(x-2)

Expand and transpose

y = (-1/2)x + 1 + 3

y = (-1/2)x + 4  ....................(1)

We choose the second equation b. x+2y=8 and convert to slope-intercept form:

2y=-x+8

y = (-1/2)x + 4, which is exactly equation (1)

So

b. x+2y=8 is the correct answer.

Answer:

b. x + 2y = 8

Step-by-step explanation:

Answer:

a) P(x=3)=0.089

b) P(x≥3)=0.938

c) 1.5 arrivals

Step-by-step explanation:

Let t be the time (in hours), then random variable X is the number of people arriving for treatment at an emergency room.

The variable X is modeled by a Poisson process with a rate parameter of λ=6.

The probability of exactly k arrivals in a particular hour can be written as:

[tex]P(x=k)=\lambda^{k} \cdot e^{-\lambda}/k!\\\\P(x=k)=6^k\cdot e^{-6}/k![/tex]

a) The probability that exactly 3 arrivals occur during a particular hour is:

[tex]P(x=3)=6^{3} \cdot e^{-6}/3!=216*0.0025/6=0.089\\\\[/tex]

b) The probability that at least 3 people arrive during a particular hour is:

[tex]P(x\geq3)=1-[P(x=0)+P(x=1)+P(x=2)]\\\\\\P(0)=6^{0} \cdot e^{-6}/0!=1*0.0025/1=0.002\\\\P(1)=6^{1} \cdot e^{-6}/1!=6*0.0025/1=0.015\\\\P(2)=6^{2} \cdot e^{-6}/2!=36*0.0025/2=0.045\\\\\\P(x\geq3)=1-[0.002+0.015+0.045]=1-0.062=0.938[/tex]

c) In this case, t=0.25, so we recalculate the parameter as:

[tex]\lambda =r\cdot t=6\;h^{-1}\cdot 0.25 h=1.5[/tex]

The expected value for a Poisson distribution is equal to its parameter λ, so in this case we expect 1.5 arrivals in a period of 15 minutes.

[tex]E(x)=\lambda=1.5[/tex]

Answer:

Option C.

Step-by-step explanation:

This is a function because all of the numbers have a partner, and none of them have more than one.

                                    Example of Not a Function

Function                                Not a Function

-4 to 5                                       -4 to 5                             <

9 to 7                                       -4 to 3                              <

13 to 3                                       13 to 3                              ^

-7 to 5                                        9 to 7                               ^

                                                 -7 to 5                               ^

                                           Not a Function because of this

Answer:

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

Step-by-step explanation:

If we are given that an object is thrown with an initial velocity of say, v1 m / s at a height of h meters, at an angle of theta ( θ ), these parametric equations would be in the following format -

x = ( 30 cos 20° )( time ),

y = - 4.9t^2 + ( 30 cos 20° )( time ) + 2

To determine " ( 30 cos 20° )( time ) " you would do the following calculations -

( x = 30 * 0.93... = ( About ) 28.01t

This represents our horizontal distance, respectively the vertical distance should be the following -

y = 30 * 0.34 - 4.9t^2,

( y = ( About ) 10.26t - 4.9t^2 + 2

In other words, our solution should be,

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

These are are parametric equations

Answer:

D.  4

Step-by-step explanation:

If he leaves the science assignments for the next day, he will spend zero hours on science assignments.  This means that y is equal to 0.  Plug this into the given equation and solve for x.

2x + y = 8

2x + 0 = 8

2x = 8

x = 4

Gerald can complete 4 math assignments.

Answer: 12/25

Steps:

1. Turn 0.48 into 48/100

2. Divide the numerator and denominator of 48/100 by 4, which ends up as 12/25.

0.48 as a fraction is 48/100

We can simplify this fraction.

48÷2/100÷2 → 24/50

24÷2/50÷2 → 12/25

Therefore, the answer is A.

Best of Luck!

Answer:

The expected value of this raffle if you buy 1​ ticket is $0.41.

Step-by-step explanation:

The expected value of the raffle if we buy one ticket is the sum of the prizes multiplied by each of its probabilities.

This can be written as:

[tex]E(X)=\sum p_iX_i[/tex]

For example, the first prize is $800 and we have only 1 prize, that divided by the number of tickets gives us a probability of 1/5000.

If we do this with all the prizes, we can calculate the expected value of a ticket.

[tex]E(X)=\sum p_iX_i\\\\\\E(X)=\dfrac{1\cdot800+3\cdot200+5\cdot50+20\cdot20}{5000}\\\\\\E(X)=\dfrac{800+600+250+400}{5000}=\dfrac{2050}{5000}=0.41[/tex]

Answer:

x = 0,00375 mm

Step-by-step explanation:

a) El factor de ampliación es 400/1   es decir el tamaño real se verá ampliado 400 veces mediante el uso del microscopio

b) De acuerdo a lo establecido en la respuesta a la pregunta referida en a (anterior) podemos establecer una regla de tres, según:

Si al microscopio el tamaño de la célula es 1,5 mm, cual será el tamaño verdadero ( que es reducido 400 en relación al que veo en el microscopio)

Es decir     1,5 mm      ⇒    400

                    x (mm)    ⇒       1 (tamaño real de la célula)

Entonces

x  =  1,5 /400

x = 0,00375 mm

Answer:

The solution is the point where the lines intersect.

The answer is (-3 , -8)

Answer:

a) 32.04% probability that overbooking occurs.

b) 40.79% probability that the flight has empty seats.

Step-by-step explanation:

For each booked passenger, there are only two possible outcomes. Either they arrive for the flight, or they do not arrive. The probability of a passenger arriving is independent of other passengers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [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]

And p is the probability of X happening.

Our variable of interest are the 8 reservations that went for the passengers with a 48% probability of arriving.

This means that [tex]n = 8, p = 0.48[/tex]

A) Find the probability that overbooking occurs.

12 seats, 8 of which are already occupied. So overbooking occurs if more than 4 of the reservated arrive.

[tex]P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)[/tex]

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

[tex]P(X = 5) = C_{8,5}.(0.48)^{5}.(0.52)^{3} = 0.2006[/tex]

[tex]P(X = 6) = C_{8,6}.(0.48)^{6}.(0.52)^{2} = 0.0926[/tex]

[tex]P(X = 7) = C_{8,7}.(0.48)^{7}.(0.52)^{7} = 0.0244[/tex]

[tex]P(X = 8) = C_{8,5}.(0.48)^{8}.(0.52)^{0} = 0.0028[/tex]

[tex]P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.2006 + 0.0926 + 0.0244 + 0.0028 = 0.3204[/tex]

32.04% probability that overbooking occurs.

B) Find the probability that the flight has empty seats.

Less than 4 of the booked passengers arrive.

To make it easier, i will use

[tex]P(X < 4) = 1 - (P(X = 4) + P(X > 4))[/tex]

From a), P(X > 4) = 0.3204

[tex]P(X = 4) = C_{8,4}.(0.48)^{4}.(0.52)^{4} = 0.2717[/tex]

[tex]P(X < 4) = 1 - (P(X = 4) + P(X > 4)) = 1 - (0.2717 + 0.3204) = 1 - 0.5921 = 0.4079[/tex]

40.79% probability that the flight has empty seats.

Answer:

10m

Step-by-step explanation:

area = 1/2 base times height

x=height

3x=base

so

150=1/2(3x^2)

300=3x^2

100=x^2

10=x

so the height is 10 and the base is 30

Answer:

h = 10

Step-by-step explanation:

Hiiiiiii

Answer:

x = y = 26 cm; z = 13 cm

Step-by-step explanation:

We can calculate the dimensions of the square base as

∛(2·8788) = 26 cm

the height of the box will be half of 26/2 which is 13 cm.

x = y = 26 cm; z = 13 cm

then the minimum area for the given volume can be calculated using what we call Lagrange multipliers, this makes it easier

area = xy +2(xz +yz)

But we were given the volume as 8788

Now we will make the partial derivatives of L to be in respect to the cordinates x, y, z, as well as λ to be equal to zero, then

L = xy +2(xz +yz) +λ(xyz -8788)

For x: we have

y+2z +λyz=0

For y we have

y: x +2z +λxz=0

For z we have 2x+2y +λxy=0............eqn(*)

For we have xyz -8788=0

If we simplify the partial derivative equation of y and x above then we have

λ = (y +2z)/(yz).

= 1/z +2/y............eqn(1)

λ = (x +2z)/(xz)

= 1/z +2/x.............eqn(2)

Set eqn(1 and 2) to equate we have

1/z +2/y = 1/z +2/x

x = y

From eqn(*) we can get z

λ = (2x +2y)/(xy) = 2/y +2/x

If we simplify we have

1/z +2y = 2/x +2/y

Then z = x/2

26/2 =13

Therefore,

x = y = 2z = ∛(2·8788)

X= 26

y = 26 cm

z = 13 cm

Answer:

D

Step-by-step explanation:

Answer:

Question 2

Step-by-step explanation:

2) The time when she woke up was -  3° C

During nature walk, temperature got 3° C warmer than when she woke up.

So, temperature during nature walk = - 3 + 3 = 0° C

Answer:

(a) The probability that all the hotel's rooms will be rented is 0.1587.

(b) The probability that 50 or more rooms will not be rented is 0.2514.

Step-by-step explanation:

We are given that the Broadmoor Hotel in Colorado Springs contains 700 rooms and is on the 2004 Gold List.

Suppose Broadmoor's marketing group forecasts a mean demand of 670 rooms for the coming weekend. Assume that demand for the upcoming weekend is normally distributed with a standard deviation of 30.

Let X = demand for rooms in the hotel

So, X ~ Normal([tex]\mu=670,\sigma^{2} =30^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                           Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = mean demand for the rooms = 670

            [tex]\sigma[/tex] = standard deviation = 30

(a) The probability that all the hotel's rooms will be rented means that the demand is at least 700 = P(X [tex]\geq[/tex] 700)

          P(X [tex]\geq[/tex] 700) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{700-670}{30}[/tex] ) = P(Z [tex]\geq[/tex] 1) = 1 - P(Z < 1)

                                                             = 1 - 0.8413 = 0.1587

The above probability is calculated by looking at the value of x = 1 in the z table which has an area of 0.8413.

(b) The probability that 50 or more rooms will not be rented is given by = P(X [tex]\leq[/tex] 650)

         P(X [tex]\leq[/tex] 650) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{650-670}{30}[/tex] ) = P(Z [tex]\leq[/tex] -0.67) = 1 - P(Z < 0.67)

                                                             = 1 - 0.7486 = 0.2514

The above probability is calculated by looking at the value of x = 0.67 in the z table which has an area of 0.7486.