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
podryga [215]
3 years ago
10

Out of 28 students in a class, 7 have blue eyes. What is the probability a student picked at

Mathematics
2 answers:
mars1129 [50]3 years ago
7 0

Answer:

3/4

Step-by-step explanation:

7/28 have blue eyes

So 28-7=21

21/28 don't have blue eyes

21/28 simplified=3/4

nadya68 [22]3 years ago
4 0

The answer would be 3/5.

You might be interested in
Before Nina bought groceries on April 22, she had a balance of $487.25 in her checking account. Nina wrote her transactions in h
Amiraneli [1.4K]

Answer: 429.75

Step-by-step explanation: i did it before

6 0
3 years ago
Read 2 more answers
Alan washed 60 plates in 30 minutes. It took him 3 minutes to wash 6 plates. Is the number of plates washed in 3 minutes proport
timofeeve [1]

Answer:

Yes

Step-by-step explanation:

The numbers both multiplied by 10

3 -6

30-60

7 0
3 years ago
Jacob and Sarah are saving money to go on a trip. They need at least $2000 in order to go. Jacob mows lawns and Sarah walks dogs
Paladinen [302]

Answer:

20x+10y\ge 2000

Step-by-step explanation:

Let x represent the number of lawns Jacob mowed and y represent the number of dogs Sarah walked.

Jacob charges $20 each time he mows a lawn, then he earns $20x for x lawns mowed.

Sarah charges $10 each time she walks a dog, then she earns $10y for y dogs walked.

In total, they will earn $(20x + 10y). They need at least $2000 in order to go on a trip, then

20x+10y\ge 2000

Note that x\ge 0,\ y\ge 0.

7 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
Mark's age, x, is 6 times his age 2 years ago.
miv72 [106K]
A, X=6(x-2) as his age take away 2 (his age two years ago) then multiplied by 6 will equal his current age.
4 0
3 years ago
Read 2 more answers
Other questions:
  • Martin used some apples to make muffins. Omar used some apples to make applesauce. Omar used 5 fewer than half as many apple as
    13·2 answers
  • 7x-3y=21<br><br> Find the intercept
    8·1 answer
  • Solve each quadratic equation by factoring and using the zero product property.
    12·2 answers
  • If the third term in a geometric sequence is 20, and the fifth term in the
    15·1 answer
  • I need help so I can pass this plz ASAP
    13·1 answer
  • I need someone to do this pls! Will give brainiest!
    6·1 answer
  • 10 POINTS PLZZZ!!!
    15·1 answer
  • a rectangular garden is 3 times as long as it is wide .the perimeter is 32m . find the dimensions of the garden​
    13·2 answers
  • Complete the following tasks for this equation: b/7.8= -2.15.
    12·1 answer
  • the life expectancy of a circulating coin is 30 years. the life expectancy of a circulating dollar bill is only 1/20 as long. fi
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!