1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Katena32 [7]
3 years ago
12

Eight kids can go to the waterpark for $60 how many students go for $195 explain how you know

Mathematics
1 answer:
cupoosta [38]3 years ago
6 0
So $60/8= $7.5 which means each ticket is $7.5
$195/$7.5= 26
That means that 26 students can buy tickets with $195
You might be interested in
Eric has 4 pieces of clay he cut out each piece of clay into thirds how many 1/3 size pieces of clay does Eric have
Ivahew [28]

Answer:

ERIC HAD 12 PIECES OF CLAY LEFT

Step-by-step explanation:

1/3 + 4 = 12

4 0
3 years ago
An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating co
levacccp [35]

Answer:

1. The 95% confidence interval would be given by (24.8190;27.8010)  

2. 4.7048 \leq \sigma^2 \leq 16.1961

3. t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871    

t_{crit}=1.753

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

4. t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306

\chi^2 =24.9958

Since our calculated value is less than the critical value we don't hav enough evidence to reject the null hypothesis at the significance level provided.

Step-by-step explanation:

Previous concepts

\bar X=26.31 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=2.8 represent the sample standard deviation

n=16 represent the sample size

Part 1

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

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

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=16-1=15

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that t_{\alpha/2}=2.13

Now we have everything in order to replace into formula (1):

26.31-2.13\frac{2.8}{\sqrt{16}}=24.819    

26.31+2.13\frac{2.8}{\sqrt{16}}=27.801

So on this case the 95% confidence interval would be given by (24.8190;27.8010)  

Part 2

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=16-1=15

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,15)" "=CHISQ.INV(0.95,15)". so for this case the critical values are:

\chi^2_{\alpha/2}=24.996

\chi^2_{1- \alpha/2}=7.261

And replacing into the formula for the interval we got:

\frac{(15)(2.8)^2}{24.996} \leq \sigma^2 \leq \frac{(15)(2.8)^2}{7.261}

4.7048 \leq \sigma^2 \leq 16.1961

Part 3

We need to conduct a hypothesis in order to determine if actual mean operating cost is at most 25 cents per mile , the system of hypothesis would be:    

Null hypothesis:\mu \leq 25      

Alternative hypothesis:\mu > 25      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

Calculate the statistic      

We can replace in formula (1) the info given like this:      

t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871    

Critical value  

On this case we need a critical value on th t distribution with 15 degrees of freedom that accumulates 0.05 of th area on the right and 0.95 of the area on the left. We can calculate this value with the following excel code:"=T.INV(0.95,15)" and we got t_{crit}=1.753

Conclusion      

Since our calculated valu it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

Part 4

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 2.3, so the system of hypothesis are:

H0: \sigma \leq 2.3

H1: \sigma >2.3

In order to check the hypothesis we need to calculate the statistic given by the following formula:

t=(n-1) [\frac{s}{\sigma_o}]^2

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 15 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,15)". And our critical value would be \chi^2 =24.9958

Since our calculated value is less than the critical value we FAIL to reject the null hypothesis.

7 0
3 years ago
Solve this please: a=46c solve for c
Firlakuza [10]
a = 46c\\c = \frac{a}{46}
7 0
3 years ago
63/40 divided by 70/44
const2013 [10]

Answer:

0.90

Step-by-step explanation:

I entered it onto my calculator, but if you'd like an explanation as to how just ask.

3 0
3 years ago
Read 2 more answers
The senior classes at Snellville High and General High planned separate trips to the water park. The senior class at Snellville
Aleonysh [2.5K]

Answer:

133

Step-by-step explanation:

Snellville

12v+14b = 796

General

14v+12b =738

We need to solve for v and b where v is how many on a van and b is how many on a bus

12v+14b = 796

14v+12b =738

Divide each equation by 2

6v+7b =398

7v+6b =369

Multiply the first equation by 7 and the second equation by -6

7(6v+7b) =7(398)  

42v +49b =2786

-6(7v+6b) =369*-6

-42v -36b = -2214

Add these equations together

42v +49b =2786

-42v -36b = -2214

-------------------------------

     13b = 572

Divide each side by 13

13b/13 = 572/13

b= 44

Now find v

6v+7b =398

6v +7(44)=398

6v +308 = 398

Subtract 308 from each side

6v = 90

Divide by 6

6v/6 = 90/6

v = 15

We want 2 b and 3 v

2b+3v = 2(44)+3(15) = 88+45=133

5 0
3 years ago
Read 2 more answers
Other questions:
  • 32 is the sum of a number z and 9
    6·2 answers
  • The displacement, d, in millimeters of a tuning fork as a function of time, t, in seconds can be modeled with the equation d = 0
    13·1 answer
  • What is the answer to that
    15·1 answer
  • Five cards are randomly selected without replacement from a standard deck of 52 playing cards. what is the probability of gettin
    14·1 answer
  • A bag contains white marbles and green marbles, 62 in total. The number of white marbles is 3 less than 4 times the number of gr
    11·1 answer
  • Four-thirds times the sum of a number and 8 is 24. What is the number?
    7·2 answers
  • Gym Membership Reach Fitness is offering yearly memberships for a flat fee of $900. However, an individual can choose to work ou
    15·1 answer
  • First to answer correctly gets brainlest no files or your answer is reported
    14·1 answer
  • WHAT IS THE MATHEMATICS OF-3 1/2
    15·1 answer
  • Two numbers differ by 3. Their product<br> is 28. Find the numbers.
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!