The average mark of candidates in an aptitude test was 128.5 with a standard deviation of 8.2.Three scores extracted from the test are 148,102,152.What is the average of the extracted scores that are outliers

Answer:

Check Explanation

Step-by-step explanation:

According to the Central limit theorem, the population mean (μ) is approximately equal to the mean of sampling distribution (μₓ).

And the standard deviation of the sampling distribution (σₓ) is related to the population standard deviation (σ) through

Standard deviation of the sampling distribution = (Population standard deviation)/(√N)

where N = Sample size

σₓ = (σ/√N)

So, population mean (μ) = Mean of sampling distribution (μₓ)

Population Standard deviation = (Standard deviation of the sampling distribution) × √N

= σ × √N

A) The expected value of a given distribution is simply equal to the mean of that distribution.

Hence, the expected value of random variable Y thay varies with different samples is given as

E(Y) = Mean of sampling distribution = μₓ

But μₓ = μ

Hence, E(Y) = μ (Proved)

B) Var (Y) is given as the square of the random distribution's standard deviation.

Var (Y) = (standard deviation of the sampling distribution)²

= (σ/√N)²

= (σ²/N) (Proved)

Hope this Helps!!!

Answer:

a) The height decreases at a rate of [tex]\frac{7}{12}[/tex] ft/sec.

b) The area increases at a rate of [tex]\frac{527}{24}[/tex] ft^2/sec

c) The angle is increasing at a rate of [tex]\frac{1}{12}[/tex] rad/sec

Step-by-step explanation:

Attached you will find a sketch of the situation. The ladder forms a triangle of base b and height h with the house. The key to any type of problem is to identify the formula we want to differentiate, by having in mind the rules of differentiation.

a) Using pythagorean theorem, we have that [tex] 25^2 = h^2+b^2[/tex]. From here, we have that

[tex]h^2 = 25^2-b^2[/tex]

if we differentiate with respecto to t (t is time), by implicit differentiation we get

[tex]2h \frac{dh}{dt} = -2b\frac{db}{dt}[/tex]

Then,

[tex]\frac{dh}{dt} = -\frac{b}{h}\frac{db}{dt}[/tex].

We are told that the base is increasing at a rate of 2 ft/s (that is the value of db/dt). Using the pythagorean theorem, when b = 7, then h = 24. So,

[tex]\frac{dh}{dt} = -\frac{2\cdot 7}{24}= \frac{-7}{12}[/tex]

b) The area of the triangle is given by

[tex]A = \frac{1}{2}bh[/tex]

By differentiating with respect to t, using the product formula we get

[tex] \frac{dA}{dt} = \frac{1}{2} (\frac{db}{dt}h+b\frac{dh}{dt})[/tex]

when b=7,  we know that h=24 and dh/dt = -1/12. Then

[tex]\frac{dA}{dt} = \frac{1}{2}(2\cdot 24- 7\frac{7}{12}) = \frac{527}{24}[/tex]

c) Based on the drawing, we have that

[tex]\sin(\theta)= \frac{b}{25}[/tex]

If we differentiate with respect of t, and recalling that the derivative of sine is cosine, we get

[tex] \cos(\theta)\frac{d\theta}{dt}=\frac{1}{25}\frac{db}{dt}[/tex] or, by replacing the value of db/dt

[tex]\frac{d\theta}{dt}=\frac{2}{25\cos(\theta)}[/tex]

when b = 7, we have that h = 24, then [tex]\cos(\theta) = \frac{24}{25}[/tex], then

[tex]\frac{d\theta}{dt} = \frac{2}{25\frac{24}{25}} = \frac{2}{24} = \frac{1}{12}[/tex]

Answer:

[tex]z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625[/tex]    

The p value for this case would be given by:

[tex]p_v =P(z<-1.625)=0.052[/tex]  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

Step-by-step explanation:

Information provided

[tex]\bar X=19.87[/tex] represent the sample mean

[tex]\sigma=0.4[/tex] represent the population deviation

[tex]n=25[/tex] sample size  

[tex]\mu_o =68[/tex] represent the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

z would represent the statistic

[tex]p_v[/tex] represent the p value

Hypothesis to test

We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 20[/tex]  

Alternative hypothesis:[tex]\mu < 20[/tex]  

The statistic is given by:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex]  (1)  

Replacing the info given we got:

[tex]z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625[/tex]    

The p value for this case would be given by:

[tex]p_v =P(z<-1.625)=0.052[/tex]  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: Impact force needed for Sap Stone tires to blow out. (foot-pounds)

n= 29, S= 1358 foot-pound

a)

Soap Stone claims that "The tires will blow out at an average pressure of μ= 26000 foot-pounds with a standard deviation of σ= 1020 foot-pounds.

According to the consumer's complaint, the variability of the blown out forces is greater than the value determined by the company.

I)

Then the parameter of interest is the population variance (or population standard deviation) and to test the consumer's complaint you have to conduct a Chi-Square test for σ².

σ²= (1020)²= 1040400 foot-pounds²

H₀: σ² ≤ 1040400

H₁: σ² > 1040400

α: 0.01

II)

[tex]X^2= \frac{(n-1)S^2}{Sigma^2} ~~X^2_{n-1}[/tex]

[tex]X^2_{H_0}= \frac{(n-1)S^2}{Sigma^2}= \frac{(29-1)*(1358)^2}{1040400} = 49.63[/tex]

III)

This test is one-tailed to the right and so is the p-value. This distribution has n-1= 29-1= 28 degrees of freedom, so you can calculate the p-value as:

P(X²₂₈≥49.63)= 1 - P(X²₂₈<49.63)= 1 - 0.99289= 0.00711

⇒ The p-value is less than the significance level so the test is significant at 1%. You can conclude that the population variance of the blowout forces is less than 1040400 foot-pounds², at the same level the population standard deviation of the blow out forces is less than 1020 foot-pounds.

b)

99% CI for the variance. Using the X² statistic you can calculate it as:

[tex][\frac{(n-1)S^2}{X^2_{n-1;1-\alpha /2}} ;\frac{(n-1)S^2}{X^2_{n-1;\alpha /2}} ][/tex]

[tex]X^2_{n-1;\alpha /2}= X^2_{28; 0.005}= 13.121[/tex]

[tex]X^2_{n-1;1-\alpha /2}= X^2_{28; 0.995}= 49.588[/tex]

[tex][\frac{28*(1358)^2}{49.588} ;\frac{28*(1358)^2}{13.121} ][/tex]

[1041312.253; 3935415.898] foot-pounds²

I hope this helps!

Answer:

about 11.11%, or 4/36

Step-by-step explanation:

Answer:

1/9

Step-by-step explanation:

There are 4 possible ways to get a sum of 5 out of total ways of 6*6= 36

So the probability of getting sum of 5 is 4/36= 1/9

Answer:

Step-by-step explanation: gradient of perpendicular line is

-2

y+1/x+3=-2

y+1=-2x-6

y=-2x-7

Answer:

B and C

Step-by-step explanation:

We need to find which one of these numbers is a perfect square. We know 7 and 13 can be eliminated since they're prime numbers so they can't be perfect squares. B and C are both perfect squares so that's the answer.

Answer:

B and C

Step-by-step explanation:

7 = 7

1.96 = [tex]\frac{196}{100}[/tex]

0.04 = [tex]\frac{4}{100}[/tex]

13 = 13

Answer: 0.0825

Step-by-step explanation:

8.25% to decimal is simply divide 8.25 by 100

8.25/100= 0.0825

The required solution after convert it into decimal is,

⇒ 0.0825

We have to given that,

Convert 8.25% into a decimal.

We can change it as,

⇒ 8.25%

Since, 1% = 1/100

Hence,

⇒ 8.25/100

⇒ 0.0825

Therefore, After convert it into decimal number solution is,

⇒ 0.0825

Learn more about the percent visit:

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#SPJ6

Answer:

T = £85.87

the cost of the petrol the car used in November is £85.87

Step-by-step explanation:

Given;

In November Hillary drove 580 miles in her car;

Distance travelled d = 580 miles

the car travelled 33.5 miles for each gallon of petrol used;

Fuel consumption rate r = 33.5 miles per gallon

Number of gallons N consumed by the car is;

N = distance travelled/fuel consumption rate

N = d/r = 580/33.5 = 17.3134 gallons

Given that;

Petrol cost £1.09 per litre

Cost per litre c = £1.09

1 gallon = 4.55 litres

Converting the amount of fuel used to litres;

N = 17.3134 gallons × 4.55 litres per gallon

N = 78.77612 litres

The total cost T = amount of fuel consumption N × fuel cost per litre c

T = N × c

T = 78.77612 litres × £1.09 per litre

T = £85.87

the cost of the petrol the car used in November is £85.87

Answer:

$25

Step-by-step explanation:

You do 1.25. Y 20

Answer:

C

Step-by-step explanation:

[tex]\frac{6}{13}[/tex]÷[tex]\frac{6}{12}[/tex]  can also be [tex]\frac{6}{13}[/tex]×[tex]\frac{12}{3}[/tex] and [tex]\frac{6}{13}[/tex]×[tex]\frac{12}{3}[/tex]=13/12

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

\frac{6}{13}136 ÷\frac{6}{12}126  can also be \frac{6}{13}136 ×\frac{12}{3}312 and \frac{6}{13}136 ×\frac{12}{3}312 =13/12

Answer:

2

Step-by-step explanation:

1+1 = 2

Hope this helps;)

Answer:

1 + 1 =2

Step-by-step explanation:

Hi, how are you?

Do u want to be my friend?

Also, I hope this answered your question?

Answer:

   {-7, -4, -1, 3, 7}

Step-by-step explanation:

The range is the list of y-coordinates of the points:

  range = {-7, -4, -1, 3, 7}

Answer:

56

Step-by-step explanation:

simple ratio: 7/95 = x/760

first, calculate 7/95 which is 0.0736

next, multiply that by 760, which would be 56

Answer:

The probability that when 12 adults are randomly selected, 3 or fewer are in excellent health is 0.2946

Step-by-step explanation:

BIONOMIAL DISTRIBUTION

pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k

where

k = number of successes in trials    

n = is the number of independent trials    

p = probability of success on each trial

I.

mean = np

where

n = total number of repetitions experiment is excueted

p = success probability

mean = 12 * 0.37

= 4.44

II.

variance = npq

where

n = total number of repetitions experiment is excueted

p = success probability

q = failure probability

variance = 12 * 0.37 * 0.63

= 2.7972

III.

standard deviation = sqrt( variance ) = sqrt(2.7972)

=1.6725

the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health

P( X < = 3) = P(X=3) + P(X=2) + P(X=1) + P(X=0)    

= ( 12 3 ) * 0.37^3 * ( 1- 0.37 ) ^9 + ( 12 2 ) * 0.37^2 * ( 1- 0.37 ) ^10 + ( 12 1 ) * 0.37^1 * ( 1- 0.37 ) ^11 + ( 12 0 ) * 0.37^0 * ( 1- 0.37 ) ^12    

= 0.2947

Answer:

0.30

Step-by-step explanation:

Data provided in the question

Uniform density function for a friend = x minutes late

The Friend is at least 21 minutes late

Based on the above information, the probability that the friend is at least 21 minutes late is

[tex]= \frac{Total\ minutes - minimum\ minutes}{Total\ minutes}[/tex]

[tex]= \frac{30 - 21}{30}[/tex]

= 0.30

Based on the above formula we can easily find out the probability for the friend who is at least 21 minutes late

The probability of the friend to be at least 21 minutes late for the uniform density function shown in the graph is 0.30.

Probability of an event is the ratio of number of favorable outcome to the total number of outcome of that event.

The graph is attached below shows the uniform density function for a friend who is x minutes late.

The friend is at least 21 minutes late. This means that the friend is 21 minutes late or more than it. 21 or more minutes goes from 21 to 30. Thus, the difference is,

[tex]d=30-21\\d=9[/tex]

The density of the graph is 1/30. The probability will be equal to the area under the curve.

In this, the length of the rectangle will be 9 and width will be 1/30 for the probability of at least 21 minutes late. The probability is,

[tex]P=9\times\dfrac{1}{30}\\P=0.30[/tex]

Thus, the probability of the friend to be at least 21 minutes late for the uniform density function shown in the graph is 0.30.

Learn more about the probability here;

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Answer:

The differential equation which describes the mixing process is [tex]\frac{dc_{salt,out}}{dt} + \frac{2}{125}\cdot c_{salt,out} = \frac{71}{100000}[/tex].

Step-by-step explanation:

The mixing process within the tank is modelled after the Principle of Mass Conservation, which states that:

[tex]\dot m_{salt,in} - \dot m_{salt,out} = \frac{dm_{tank}}{dt}[/tex]

Physically speaking, mass flow of salt is equal to the product of volume flow of water and salt concentration. Then:

[tex]\dot V_{water, in, 1}\cdot c_{salt, in,1} + \dot V_{water, in, 2} \cdot c_{salt,in, 2} - \dot V_{water, out}\cdot c_{salt, out} = V_{tank}\cdot \frac{dc_{salt,out}}{dt}[/tex]

Given that [tex]\dot V_{water, in, 1} = 9\,\frac{L}{min}[/tex], [tex]\dot V_{water, in, 2} = 7\,\frac{L}{min}[/tex], [tex]c_{salt,in,1} = 0.04\,\frac{kg}{L}[/tex], [tex]c_{salt, in, 2} = 0.05\,\frac{kg}{L}[/tex], [tex]\dot V_{water, out} = 16\,\frac{L}{min}[/tex] and [tex]V_{tank} = 1000\,L[/tex], the differential equation that describes the system is:

[tex]0.71 - 16\cdot c_{salt,out} = 1000\cdot \frac{dc_{salt,out}}{dt}[/tex]

[tex]1000\cdot \frac{dc_{salt, out}}{dt} + 16\cdot c_{salt, out} = 0.71[/tex]

[tex]\frac{dc_{salt,out}}{dt} + \frac{2}{125}\cdot c_{salt,out} = \frac{71}{100000}[/tex]

Answer:

V =2744 cm^3

Step-by-step explanation:

The volume of a cube is given by

V = s^3 where s is the side length

V = 14^3

V =2744 cm^3

Answer: 2,744 cm³

Step-by-step explanation: Since the length, width, and height of a cube are all equal, we can find the volume of a cube by multiplying side × side × side.

So we can find the volume of a cube using the formula s³.

Notice that we have a side length of 14cm.

So plugging into the formula, we have (14 cm)³ or

(14 cm)(14 cm)(14 cm) which is 2,744 cm³.

So the volume of the cube is 2,744 cm³.

Answer:

The quadratic formula is x= -b±√b²-4ac all over 2a

Step-by-step explanation:

The -b±√b²-4ac is part of the numerator of the fraction. And b²-4ac is all under the radical sign. The denominator is 2a so you divide -b±√b²-4ac by 2a to get the solutions of the quadratic function. Hope this helps!!

Answer:

The data in shelter A show more spread than the data in shelter B.

Step-by-step Explanation:

Judging from the data set given for shelter A and shelter B in the diagram of the box plots attached below, shelter A has a data set that ranges from 8 pounds to 30 pounds, while shelter B ranges from e can infer that the data in shelter A has more spread than the data in shelter B ranges from 10 to 28.

Taking into consideration the range of the data set of each shelter, shelter A obviously has a larger range compared to shelter B. This implies that, the data in shelter A show more spread than the data in shelter B.

Answer:

A

Step-by-step explanation:

Answer:

product is right

Step-by-step explanation:

Complete  Question

The complete question is shown on the first uploaded image

Answer:

 a

    b can be written as a linear combination of  [tex]a_1 \ and \ a_2[/tex]

b

   The values of   [tex]x_1 = 4 \ and \ x_2 = 2[/tex]

Step-by-step explanation:

From the question we are told that

      [tex]x_1 a_1 +x_2 a_2 = b[/tex]

Where [tex]a_ 1 = (4, 5,-4)[/tex],  [tex]a_2 = (-4 , 3, 3)[/tex] and  [tex]b = (8,26 , -10)[/tex]

   So  

      [tex]x_1 ( 4, 5,-4) + x_2 (-4 , 3, 3) = (8,26 , -10)[/tex]

     [tex]4x_1, 5x_1,-4x_1 + -4x_2 , 3x_2, 3x_2 = (8,26 , -10)[/tex]

=>   [tex]4x_1 -4x_2 =8[/tex]

     [tex]x_1 -x_2 =2 ---(1)[/tex]

=>   [tex]5x_1 + 3x_2 = 26 --- (2)[/tex]

=>   [tex]-4x_1 + 3x_2 = -10 ---(3)[/tex]

Now multiplying equation 1 by 3 and adding the product to equation 2

          [tex].\ \ \ 3x_1 -3x_2 = 6\\+ \ \ 5x_1 + 3x_2 = 26 \\=> \ \ \ 8x_1 = 32[/tex]

=>        [tex]x_1 = 4[/tex]

 substituting [tex]x_1[/tex] into equation 1

        [tex]4 - x_2 =2[/tex]

        [tex]x_2 =2[/tex]

Now to test substitute [tex]x_1 \ and \ x_2[/tex] into equation 3

       [tex]-4(4) + 3(2) = -10[/tex]

      [tex]-10 = -10[/tex]

Since LHS = RHS then there exist values [tex]x_1 = 4 \ and \ x_2 = 2[/tex]  such that

       [tex]x_1 a_1 +x_2 a_2 = b[/tex]

Hence b can be written as a linear combination of  [tex]a_1 \ and \ a_2[/tex]

Answer:

Step-by-step explanation:

The sine of an angle is equal to the cosine of its complement, and vice versa.

  sin ∠QOP = cos ∠ROQ

  cos ∠ROQ = sin ∠QOP

Answer:

YlThe higher the grader the more hours they can work.

Step-by-step explanation:

We can see that the higher the grader the more hours they can work ; which could mean less academic work if the work defined is that with which to earn money but if the work defined is academic it means more hours of academic work

Answer:

The number of loaves to be made daily to maximize Mr. Joaquim's average profit is 1400.

O número de pães a serem feitos diariamente para maximizar o lucro médio do Sr. Joaquim é 1400.

Step-by-step explanation:

The table gives the daily demand of bread and the average profit that corresponds to each demand.

The daily demand values recorded ranged from 1000 to 1800.

And the corresponding average profit for the demands for each daily demand provided showed a steady increase from $0.10 as at a daily demand of 1000, through $0.15, then peaking at $0.20 at a daily demand of 1400.

After this point, the average profit begins to decline again as we move further away from the peak obtained at a daily demand of 1400.

Hence, it is evident that the daily demand that corresponds to the maximum average profit is 1400.

A daily demand of 1400 is where the peak average profit of $0.20 is obtained.

Any other demand lower or higher than 1400 corresponds to a departure from the peak profit obtained at that point according to the table of daily demand and average profits provided in the question.

In portugese/Em português

A tabela fornece a demanda diária de pão e o lucro médio que corresponde a cada demanda.

Os valores diários de demanda registrados variaram de 1000 a 1800.

E o lucro médio correspondente às demandas de cada demanda diária fornecida mostrou um aumento constante de US $ 0,10, com uma demanda diária de 1000, passando por US $ 0,15, atingindo um pico de US $ 0,20 com uma demanda diária de 1400.

Após esse ponto, o lucro médio começa a declinar novamente à medida que nos afastamos do pico obtido com uma demanda diária de 1400.

Portanto, é evidente que a demanda diária que corresponde ao lucro médio máximo é de 1.400.

Uma demanda diária de 1400 é onde é obtido o lucro médio máximo de US $ 0,20.

Qualquer outra demanda menor ou maior que 1400 corresponde a um afastamento do pico de lucro obtido nesse momento, de acordo com a tabela de demanda diária e lucros médios fornecidos na pergunta.

Hope this Helps!!!

Espero que isto ajude!!!

Answer:

x= 4

Y= -1

Z= 3

step by step procedure:

let x+y+z=6, x-y+z=8, x+y-z=0 be equations 1, 2, and 3 respectively.

you are welcome bro....

Answer:

a. The Poisson approximation is good because n is large, p is small, and np < 10.

The parameter of thr Poisson distribution is:

[tex]\lambda =np\approx1.6[/tex]

b. P(r=0)=0.2019

c. P(r>1)=0.4751

d. P(r>2)=0.2167

e. P(r>3)=0.0789

Step-by-step explanation:

a. The Poisson distribution is appropiate to represent binomial events with low probability and many repetitions (small p and large n).

The approximation that the Poisson distribution does to the real model is adequate if the product np is equal or lower than 10.

In this case, n=963 and p=1/596, so we have:

[tex]np=963*(1/596)\approx1.6[/tex]

The Poisson approximation is good because n is large, p is small, and np < 10.

The parameter of thr Poisson distribution is:

[tex]\lambda =np\approx1.6[/tex]

We can calculate the probability for k events as:

[tex]P(r=k)=\dfrac{\lambda^ke^{-\lambda}}{k!}[/tex]

b. P(r=0). We use the formula above with λ=1.6 and r=0.

[tex]P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\[/tex]

c. P(r>1). In this case, is simpler to calculate the complementary probability to P(r<=1), that is the sum of P(r=0) and P(r=1).

[tex]P(r>1)=1-P(r\leq1)=1-[P(r=0)+P(r=1)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\\\P(r>1)=1-(0.2019+0.3230)=1-0.5249=0.4751[/tex]

d. P(r>2)

[tex]P(r>2)=1-P(r\leq2)=1-[P(r=0)+P(r=1)+P(r=2)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\P(2)=1.6^{2} \cdot e^{-1.6}/2!=2.56*0.2019/2=0.2584\\\\\\P(r>2)=1-(0.2019+0.3230+0.2584)=1-0.7833=0.2167[/tex]

e. P(r>3)

[tex]P(r>3)=1-P(r\leq2)=1-[P(r=0)+P(r=1)+P(r=2)+P(r=3)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\P(2)=1.6^{2} \cdot e^{-1.6}/2!=2.56*0.2019/2=0.2584\\\\P(3)=1.6^{3} \cdot e^{-1.6}/3!=4.096*0.2019/6=0.1378\\\\\\P(r>3)=1-(0.2019+0.3230+0.2584+0.1378)=1-0.9211=0.0789[/tex]

Answer:

Hey!

Your answer is 150 square meters!

Step-by-step explanation:

13*3=39 (the tilted face)

12*3=36 (the flat face)

5*3=15 (the base)

TO FIND THE TRINGLULAR AREA:

1/2 base x height...

1/2 x 5 x 12 = 30

(WE DOUBLE THIS BECAUSE THERE ARE TWO TRIANGLES)

SO 60...

ADD THESE TOGETHER...

39 + 36 + 15 + 30 + 30...

GIVES US 150 square metres

Answer:

150 square meters

Step-by-step explanation:

Answer:

200.96 units

Step-by-step explanation:

Use the formula for the volume of a cone [tex]V=\pi r^{2} \frac{h}{3}[/tex]

Plug in the values ([tex]\pi[/tex]=3.14) and multiply them all out

Answer:

≈ 201

Step-by-step explanation:

V= πr²h/3

V= 3.14*4²*12/3= 200.96 ≈ 201