Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally distributed with a standard deviation 20 minutes, answer the following questions. (A) What is the probability that a randomly selected Time interval between irruption’s is longer than 84 minutes? (B) what is the probability that a random sample of 13 time intervals between irruption‘s has a mean longer than 84 minutes? (C) what is the probability that a random sample of 20 time intervals between irruption‘s has a mean longer than 84 minutes? (D) what effect does increasing the sample size have on the probability? Provide an exclamation for this result. Choose the correct answer below. (E) what might you conclude if a random sample of 20 time intervals between irruption‘s has a mean longer than 84 minutes? Choose the best answer below. I’m not entirely certain about my answer for a bit I am completely and utterly lost on the other questions... please help.

Suppose A Geyser Has A Mean Time Between Irruptions Of 75 Minutes. If The Interval Of Time Between The

Answers

Answer 1

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal([tex]\mu=75, \sigma^{2} =20[/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] = population mean time between irruption = 75 minutes

           [tex]\sigma[/tex] = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{84-75}{20}[/tex] ) = P(Z > 0.45) = 1 - P(Z [tex]\leq[/tex] 0.45)

                                                        = 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let [tex]\bar X[/tex] = sample time intervals between the eruption

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

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

where, [tex]\mu[/tex] = population mean time between irruption = 75 minutes

           [tex]\sigma[/tex] = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P([tex]\bar X[/tex] > 84 min)

 

    P([tex]\bar X[/tex] > 84 min) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{84-75}{\frac{20}{\sqrt{13} } }[/tex] ) = P(Z > 1.62) = 1 - P(Z [tex]\leq[/tex] 1.62)

                                                        = 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let [tex]\bar X[/tex] = sample time intervals between the eruption

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

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

where, [tex]\mu[/tex] = population mean time between irruption = 75 minutes

           [tex]\sigma[/tex] = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P([tex]\bar X[/tex] > 84 min)

 

    P([tex]\bar X[/tex] > 84 min) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{84-75}{\frac{20}{\sqrt{20} } }[/tex] ) = P(Z > 2.01) = 1 - P(Z [tex]\leq[/tex] 2.01)

                                                        = 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.


Related Questions

PLS HURRY!!!! The total amount of water a tank can hold is 14 and one-half gallons. Raul wants to find out how many 1 and one-fourth-gallon buckets of water can be used to fill the tank. Which expressions could be used to represent this scenario? Select two options.. StartFraction 29 over 2 EndFraction divided by StartFraction 5 over 4 EndFraction StartFraction 29 over 2 EndFraction times StartFraction 5 over 4 EndFraction StartFraction 29 over 2 EndFraction times StartFraction 4 over 5 EndFraction StartFraction 2 over 29 EndFraction times StartFraction 5 over 4 EndFraction StartFraction 29 over 2 EndFraction divided by StartFraction 4 over 5 EndFraction

Answers

Answer:

[29 over 2 divided by 5 over 4] = 29/2 ÷ 5/4 (A)

[29 over 2 times 4 over 5] = 29/2 × 4/5 (C)

Step-by-step explanation:

The total amount of water a tank can hold is 14 and one and one-half gallons = 14½

We are to determine the number of 1/4 gallon buckets of water that can be used to fill the tank.

To find this, we would divide 14½ by 1¼ = 14½ ÷ 1¼

Convert 14½ and 1¼ to improper fraction

14½ = [(14×2)+1]/2 = 29/2

1¼ = [(1×4)+1]/4 = 5/4

14½ ÷ 1¼ =29/2 ÷ 5/4

Then we find the inverse of ¼ to convert the division to multiplication

= 29/2 × 4/5 = 116/10

= 11.6

14½ ÷ ¼ = 11.6

Therefore 11.6 gallons of ¼ buckets of water can be used to fill the tank.

But the two expressions that could be used to represent this scenario to get the answer from the options = 29/2 ÷ 5/4 (A)

[29 over 2 divided by 5 over 4]

29/2 × 4/5 (C)

[29 over 2 times 4 over 5]

Answer:

I believe it is A and C.

The mayor of a town has proposed a plan for the construction of an adjoining community. A political study took a sample of 900 voters in the town and found that 45% of the residents favored construction. Using the data, a political strategist wants to test the claim that the percentage of residents who favor construction is over 42%. Determine the P-value of the test statistic. Round your answer to four decimal places.

Answers

Answer:

P-value = 0.0367

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that the percentage of residents who favor construction is significantly over 42%.

Then, the null and alternative hypothesis are:

[tex]H_0: \pi=0.42\\\\H_a:\pi>0.42[/tex]

The sample has a size n=900.

The sample proportion is p=0.45.

 

The standard error of the proportion is:

[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.42*0.58}{900}}\\\\\\ \sigma_p=\sqrt{0.000271}=0.016[/tex]

Then, we can calculate the z-statistic as:

[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.45-0.42-0.5/900}{0.016}=\dfrac{0.029}{0.016}=1.79[/tex]

This test is a right-tailed test, so the P-value for this test is calculated as:

[tex]\text{P-value}=P(z>1.79)=0.0367[/tex]

Look at the picture for the question ........................

Answers

Answer:

53/28

Step-by-step explanation:

100 g jam

19 g strawberries53 g blackberriesSugar= 100-(19+53)= 28 g

Blackberry / sugar= 53/28

Answer:

53:28

Solution,

Amount of sugar=100-(19+53)

=100-72

=28 gram

Ratio of blackberries to sugar=53/28

=53:28

Hope this helps..

Good luck on your assignment..

i will give brainliest and 50 points pls help ASP

Answers

Answer:

Step-by-step explanation:

A trapezoid and two rectangulars are in the  opening form of the geometric shapes and you should calculate their surface areas seperately which is:

12 cm and 30 cm first rectangular and its surface area is 12 x 30 = 360 cm^2second rectangular has 30 cm and 26 cm and 30 x 26 = 780 cm^2A trapezoid surface area is = [(24+29) /2] * 25 = 662.5 cm^2total surface area = 360 + 780 + 662.5 = 1802.5

Answer:

Total surface area  = 3744 cm^2

Step-by-step explanation:

All linear measurements are in cm

Surface area of BOTH bases

Ab = 2*  (12+29)*24/2

= 984

Circumference of base

Cb = (25+12+26+29)

= 92

Height of prism

H = 30  (given)

Surface area of sides of prism

As= Cb*H

= 2760

Total Surface area of Prism

A = Ab + As

= 984 + 2760

= 3744 cm^2

Given the equation A=250(1.1)t, you can determine that the interest is compounded annually and the interest rate is 10%. Suppose the interest rate were to change to being compounded quarterly. Rewrite the equation to find the new interest rate that would keep A and P the same. What is the approximate new interest rate? Convert your answer to a percentage, round it to the nearest tenth, and enter it in the space provided, like this: 42.53%

Answers

Answer:

[tex]r \approx 2.41\%[/tex]

Step-by-step explanation:

The computation of the approximate new interest rate is shown below:

As we know that there are four quarters in a year so

The new equation is

[tex]A = 250(1 + r)^{4t}[/tex]

Now to determine the value of interest rate,i.e r, so place this to the first equation.

So,

[tex]250(1.1)^{t} = 250(1 + r)^{4t}[/tex]

[tex]1.1^{t} = (1 + r)^{4t}[/tex]

1.1 = (1 + r)^4

[tex]\sqrt[4]{1.1} = 1 + r[/tex]

[tex]r = -1 + \sqrt[4]{1.1}[/tex]

[tex]r \approx 0.0241[/tex]

[tex]r \approx 2.41\%[/tex]

We simply applied the above formula so that the interest rate could come

Could someone explain how to find square roots please?

Answers

Answer: graphing calculator!

Step-by-step explanation: if you’re looking for the square root of a # that isn’t a perfect square (ie. sqrt4, sqrt 36) then you have to use a calculator for that. however the idea behind square roots is just a # multiplied by itself to give the original #. just ask yourself “what can i multiply by itself to get the original number”. hope that helped !

Answer:

By multiplying the number by its power 2.

E.g= 4^2

How do I calculate velocity?

Answers

Answer:

v = Δs/Δt

Step-by-step explanation:

Velocity is equal to the displacement/distance (delta symbol s) over the change of time (delta symbol t).

Select all that apply. A direct variation equation is graphed. The point (6, 4) is on the graph. Which of the following points are also on the graph of the equation? (3, 2) (-9, -6) (-12, 8) (-6, -3)

Answers

Answer:

(3,2)

(-9,-6)

Step-by-step explanation:

Given that the graph is a direct variation.

The equation of variation is:

[tex]x=ky[/tex]

Since point (6, 4) is on the graph

[tex]6=4k\\\\k=\dfrac64[/tex]

Therefore, the equation connecting x and y is:

[tex]x=\dfrac64y[/tex]

[tex]\text{When y=2},x=\dfrac64 \times 2 =3\\\\\text{When y =}-6,x=\dfrac64 \times -6 =-9\\\\\text{When y =}8,x=\dfrac64 \times 8 =12\\\\\text{When y =}-3,x=\dfrac64 \times -3 =-4.5[/tex]

Therefore, the points that are also on the graph are:

(3,2) and  (-9,-6)

Answer:

(-9, -6) and (3, 2)

Step-by-step explanation:

The graph of y = 1/2 x2 + 2x + 3 is shown. What are the solutions to the equation 1/2 x2 + 2x + 3 = x + 7?

Answers

Answer:

Step-by-step explanation:

Let's organise our information :

the function is (1/2)x²+2x+3

we want to khow the value of x that gives us : (1/2)x²+2x+3=x+7

Now the trick is to write this expression as a quadratic equation with zero at one side :

(1/2)x²+2x+3=x+7 (1/2)x²+2x+3-x-7=0(1/2)x²+x-4=0

Now let's solve this equation :

a= 1/2b= 1c= -4

Δ=1²-4*(1/2)*(-4)

 = 9

So we have two solutions :

[tex]\left \{ {{y=\frac{-1-3}{2*0.5} } \atop {x=\frac{-1+3}{2*0.5} }} \right.[/tex]y= -4x= 2

So the solutions are -4 and 2

Please answer question now

Answers

Answer:

∠CED

or

∠DEC

its the same thing

Step-by-step explanation:

E is the angle

In the morning, Marco sold 12 cups of lemonade for $3. By the end of the day, he had earned $9. How many cups of lemonade did he sell in all? Kaycee wrote the proportion 12/3=9/c for this situation identifly the error and give tow ways to write the proportion correctly

Answers

Answer:

36 cups

3/12 = 9/c

Step-by-step explanation

3 ÷ 12 = .25

each cup = 25 cents

9 ÷ .25 = 36

Marco sold 36 cups of lemonade.

3/12 = 9/c

Answer:

36 cups

Step-by-step explanation:

We can set up a proportion:

12/3=c/9

Cross multiply.

12*9=3*c

108=3c

Divide both sides by 3.

c=36

The error that the proportion Kaycee set up was

cups/price=price/cups

This is not proportionate.

Possible proportions include:

cups/price=cups/price

price/cups=price/cups

cups/cups=price/price

price/price=cups/cups

algebra ...........................

Answers

Answer:

see explanation

Step-by-step explanation:

Given that y is inversely proportional to x then the equation relating them is

y = [tex]\frac{k}{x}[/tex] ← k is the constant of proportion

(a)

To find k use the condition when y = 7 , x = 9, that is

7 = [tex]\frac{k}{9}[/tex] ( multiply both sides by 9 )

63 = k

y = [tex]\frac{63}{x}[/tex] ← equation of proportion

(b)

When x = 21, then

y = [tex]\frac{63}{21}[/tex] = 3

Melvin has game and education apps on his tablet. He noticed that he has 3 game apps for every 2 education apps. Which of the following is another way to write this ratio? 1:2 2:1 2:3 3:2

3:3

Answers

the correct answer would be 3:2

Answer:

3: 2

Step-by-step explanation:

game Apps: education apps:

3: 2

In a certain city, the true probability of a baby being a boy is 0.559. Among the next eight randomly selected births, find the probability that at least one of them is a boy. (round to three decimal places)

Answers

Answer:

99.86%

Step-by-step explanation:

What we have here is a binomial distribution, let x be the baby number among 8 births.

Binomial (n = 8, p = 0.559)

We want to find, P (X ≥ 1)

P (X ≥ 1)    = 1 - P (X <1)

P (X ≥ 1)  = 1 - P (X = 0)

P (X = 0)   = 8C0 * (0.559) ^ 0 * (1 - 0.559) ^ (8-0)]

8C0 = 8! / (0! * (8-0)!) = 1

replacing we have:

P (X = 0)   = 1 * 1 * 0.0014

P (X = 0)  = 0.0014

P (X ≥ 1))  = 1 - 0.001 4

P (X ≥ 1) = 0.9986

Therefore the probability is 99.86%

helppppppppp hurryyyyyyyyyyyyyyyyyyyyyyyy

Answers

Answer:

2

Step-by-step explanation:

I would have to say it is two bc Everitt had 30% where Desery had 20%.

Randomthing2143 help me please! Range of: 1, 2, 3, 1, 5, 2, 1, 8, 9, 7?

Answers

Answer:

8

solution,

Highest score=9

lowest score=1

now,

range= highest score-lowest score

=9-1

=8

Hope this helps...

Answer:

8

Step-by-step explanation:

The range of the sequence is the largest number - the smallest number.

The largest number is 9.

The smallest number is 1.

9 - 1

= 8

The range of the sequence is 8.

Which of the following represents the ratio of the hypotenuse to the given side?

Answers

Answer:c

Step-by-step explanation:

hyp=6[tex]\sqrt{2}[/tex]

adj=6

hyp:adj=6[tex]\sqrt{2}[/tex]  :2

hyp:adj=[tex]\frac{6\sqrt{2} }{6} \\[/tex]

[tex]=\frac{\sqrt{2} }{1} \\hyp:adj=\sqrt{2} :1[/tex]

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 4.2.

Required:
Provide 90%90% and 95%95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

Answers

Answer:

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and the critical value would be [tex]t_{\alpha/2}=1.669[/tex]

And replacing we got

[tex]17.5-1.669\frac{4.2}{\sqrt{65}}=16.63[/tex]    

[tex]17.5+1.669\frac{4.2}{\sqrt{65}}=18.37[/tex]    

For the 95% confidence the critical value is [tex]t_{\alpha/2}=1.998[/tex]

[tex]17.5-1.998\frac{4.2}{\sqrt{65}}=16.46[/tex]    

[tex]17.5+1.998\frac{4.2}{\sqrt{65}}=18.54[/tex]    

Step-by-step explanation:

Information given

[tex]\bar X¿ 17.5[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s¿4.2 represent the sample standard deviation

n¿65 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

The degrees of freedom are given by:

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

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and the critical value would be [tex]t_{\alpha/2}=1.669[/tex]

And replacing we got

[tex]17.5-1.669\frac{4.2}{\sqrt{65}}=16.63[/tex]    

[tex]17.5+1.669\frac{4.2}{\sqrt{65}}=18.37[/tex]    

For the 95% confidence the critical value is [tex]t_{\alpha/2}=1.998[/tex]

[tex]17.5-1.998\frac{4.2}{\sqrt{65}}=16.46[/tex]    

[tex]17.5+1.998\frac{4.2}{\sqrt{65}}=18.54[/tex]    

A department store chain is expanding into a new market, and is considering 13 different sites on which to locate 5 stores. Assuming that each site is equally likely to be chosen, in how many ways can the sites for the new stores be selected?

Answers

Answer:

Hope it helps you :)

And i hope you understand  

For store 1, it can be placed on 16 sites. Store 2 can be placed on 15 sites( since store 1 is already on site 1). Store 3 can be placed on 14 sites and so on until store 5 which has 12 sites.

Therefore the number of way is

C= 16*15*14*13*12

c=524,160 possibilities

Evaluate. Write your answer as a fraction or whole number without exponents. 3^–3 =

Answers

Answer:

The answer is 1/27

Step-by-step explanation:

According to the rules of indices

a^-b can be written as 1/a^b

So 3^- 3 can be written as 1/3³

And

1/3³ = 1/27

Hope this helps you

Quadrilaterals WXYZ and BADC are congruent. In addition, WX ≅ DC and XY ≅ BC.

If AD = 4 cm and AB = 6 cm, what is the perimeter of WXYZ?

18 cm
20 cm
22 cm
24 cm

Answers

Answer: 20 cm

If quadrilaterals WXYZ and BADC are congruent, then their corresponding sides are congruent.

Given that

WX≅DC,

XY≅BC,

you can state that

YZ≅AB,

WZ≅AD.

If AD = 4 cm and AB = 6 cm, then WZ = 4 cm and YZ = 6 cm. Opposite rectangle sides are congruent, then XY = 4 cm and WX = 6 cm.

The perimeter of WXYZ is

P = WX + XY + YZ + WZ = 6 + 4 + 6 + 4 = 20 cm.

Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more than $55,000 per year. Gina, his colleague, believes this to be incorrect, so she randomly selects 61 employees who work in Yarmouth and records their annual salaries. Gina calculates the sample mean income to be $56,500 per year with a sample standard deviation of 3,750. Using the alternative hypothesis Ha:μ>55,000, find the test statistic t and the p-value for the appropriate hypothesis test. Round the test statistic to two decimal places and the p-value to three decimal places.

Degrees of Freedom

0.0004 0.0014 0.0024 0.0034 0.0044 0.0054 0.0064
54 3.562 3.135 2.943 2.816 2.719 2.641 2.576
55 3.558 3.132 2.941 2.814 2.717 2.640 2.574
56 3.554 3.130 2.939 2.812 2.716 2.638 2.572
57 3.550 3.127 2.937 2.810 2.714 2.636 2.571
58 3.547 3.125 2.935 2.808 2.712 2.635 2.569
59 3.544 3.122 2.933 2.806 2.711 2.633 2.568
60 3.540 3.120 2.931 2.805 2.709 2.632 2.567

Answers

Answer:

Test statistic t = 3.12

P-value = 0.001

As the P-value (0.001) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the average salary of an employee in the city of Yarmouth is is significantly greater than $55,000 per year.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the average salary of an employee in the city of Yarmouth is is significantly greater than $55,000 per year (Gina's claim).

Then, the null and alternative hypothesis are:

[tex]H_0: \mu=55000\\\\H_a:\mu> 55000[/tex]

The significance level is 0.05.

The sample has a size n=61.

The sample mean is M=56500.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=3750.

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

[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{3750}{\sqrt{61}}=480.138[/tex]

Then, we can calculate the t-statistic as:

[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{56500-55000}{480.138}=\dfrac{1500}{480.138}=3.12[/tex]

The degrees of freedom for this sample size are:

df=n-1=61-1=60

This test is a right-tailed test, with 60 degrees of freedom and t=3.12, so the P-value for this test is calculated as (using a t-table):

[tex]\text{P-value}=P(t>3.12)=0.001[/tex]

As the P-value (0.001) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the average salary of an employee in the city of Yarmouth is is significantly greater than $55,000 per year.

use the drop downs to identify the domain and range of the following relation.
{(-4, -7), (0, 6), (5, -3), (5, 2)}​

Answers

Answer:Domain is x and range is y.For ex:-4 is domain and -7 is range.

Step-by-step explanatioFeel pleasure to help u:

Domain ( -4, 5) Range ( -7, 6)

The Domain includes the numbers between the least and the greatest x-values.

The range includes the numbers between the lowest and the highest y-values.

Which two equations are the equations of vertical asymptotes of the function y = 5∕3 tan(3∕4x)?
A) x-2pi/3 and x=-2pi/3
B) x=0 and x=2pi/3
C) x=4pi/3 and x =4pi/3
D) x=0 and x=4pi/3

I did not know how to paste the pi symbol so I used the letters (pi)

Answers

Answer:

A)x=2pi/3 and x=-2pi/3

Step-by-step explanation:

The function [tex]y=\frac{5}{3}tan(\frac{3}{4}x)[/tex] has vertical asymptotes in the values where the tan(a) has vertical asymptotes.

we know that tan(a) has vertical asymptotes in [tex]a=\frac{\pi }{2}[/tex] and  [tex]a=\frac{-\pi }{2}[/tex], if we made [tex]a=\frac{3x}{4}[/tex] and solve for x, we get:

for [tex]a=\frac{\pi }{2}[/tex]

[tex]\frac{\pi }{2} =\frac{3x}{4}\\x = \frac{2\pi }{3}[/tex]

for [tex]a=\frac{-\pi }{2}[/tex]

[tex]\frac{-\pi }{2} =\frac{3x}{4}\\x = \frac{-2\pi }{3}[/tex]

Finally, the function [tex]y=\frac{5}{3}tan(\frac{3}{4}x)[/tex] has vertical asymptotes in the values x=2pi/3 and x=-2pi/3

Answer:

A

Step-by-step explanation:

It is known that 10% of the calculators shipped from a particular factory are defective. What is the probability that exactly three of five chosen calculators are defective

Answers

Answer:0.0081  or 0.81%

Step-by-step explanation:

The required probability is P(3,5,0.1)= C5 3 * p^3*q^2, where

C5 3=  5!/3/2=4*5/2=10

p is the probability that one randomly selected calculator is defective= 10%=0.1

q is the probability that one randomly selected calculator is non-defective.

q=1-p=1-0.1=0.9

So P(3,5,0.1)= 10*0.1^3*0.9^2=0.01*0.81=0.0081

Juanita wants to buy 12.5 pounds of stones to decorate her garden. The stone normally sells for $1.80 a pound. The salesman offers her a total price of $21.25. How much is she saving per pound with the salesman’s offer? Round to the nearest hundredth.

Answers

Answer:

She is saving 10 cents per pound

Step-by-step explanation:

First, we need to find how much she would pay without the sale:

12.5(1.8) = 22.5

Now, we can calculate how much she saved:

22.5 - 21.25 = 1.25

To find how much she's saving per pound, we divide the difference by the number of pounds:

1.25/12.5 = 0.1

She is saving 10 cents per pound

Answer:

A

Step-by-step explanation:

I took the test and got it right

Which of the following statements is false regarding randomization?

Answers

Answer:b

Step-by-step explanation:

what is the surface area of the prism shown to the left?

(idk the answer i guessed)

Answers

Answer:

828

Step-by-step explanation:

Missing leg of triangle:

√15²-12²= √225-144=√81=9

Triangle faces= 2*1/2*12*9= 108

Base= 20*9= 180

Front face= 20*15= 300

Back face= 20*12= 240

Total= 108+180+300+240= 828

Second choice

ALGEBRA Identify the similar triangles. Then find each measure.
FG
S
G
6
х
R
4
T
F
10
H
ARST - AFGHFG = 11
ARST - AFGHFG = 14
ARST - AFGH FG = 15
ARST- AFGH FG = 18

Answers

Answer:

fg t f h 10

Step-by-step explanation:

Apply t-test to the following data sets and compare it with the value of u=4. Identify the test and find the critical value and state the conclusion
a) 2 3 4 5 6 7 b) 1 2 2 5 5 7

Answers

Answer:

a)0.654

b)-0.34

Step-by-step explanation:

a)2 3 4 5 6 7

[tex]\mu = 4[/tex]

[tex]Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{2+3+4+5+6+7}{6}\\Mean =4.5[/tex]

Standard deviation [tex]=\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}[/tex]

Standard deviation [tex]=\sqrt{\frac{(2-4.5)^2+(3-4.5)^2+(4-4.5)^2+(5-4.5)^2+(6-4.5)^2+(7-4.5)^2}{6-1}}=1.87[/tex]

[tex]t=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}\\t=\frac{4.5-4}{\frac{1.87}{\sqrt{6}}}\\t=0.654[/tex]

Df = n-1 = 6-1 = 5

Assume α=0.05

So[tex]t_{(\alpha,df)}=t_{0.05,5}=2.57[/tex]

So, t critical = 2.57

So, t calculated < t critical

b)1 2 2 5 5 7

[tex]\mu = 4 \\Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{1+2+2+5+5+7}{6}\\Mean =3.67[/tex]

Standard deviation =[tex]\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}[/tex]

Standard deviation =[tex]\sqrt{\frac{(1-3.67)^2+(2-3.67)^2+(2-3.67)^2+(5-3.67)^2+(5-3.67)^2+(7-3.67)^2}{6-1}}=2.338[/tex]

[tex]t=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}t=\frac{3.67-4}{\frac{2.338}{\sqrt{6}}}t=-0.34[/tex]

Df = n-1 = 6-1 = 5

Assume α=0.05

So, [tex]t_{(\alpha,df)}=t_{0.05,5}=2.57[/tex]

So, t critical = 2.57

So, t calculated < t critical

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