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Maksim231197 [3]
3 years ago
8

The stem and leaf plot below shows the number of miles run by members of the cross country team. How many students ran more than

27 miles?
A. 3

B. 5

C. 8

D. 9

Mathematics
2 answers:
nata0808 [166]3 years ago
6 0

Answer:

9 students.

Step-by-step explanation:

We see that 2 students ran 5 and 6 miles. Counting we also see 3 ran  between  10 and 19 and 3 ran 22 26 and 27 miles.

Number who ran more than 27 miles =  9 students.

oksano4ka [1.4K]3 years ago
5 0

Answer:

D

Step-by-step explanation:

Counting from 28 to 44, that is more than 27 miles

28, 31, 32, 33, 35, 39, 42, 43, 44

That is 9 ran more than 27 miles

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I tried to figure it out but i can't. i need to show the work and i don't know how to do it.​
Andreas93 [3]

Integration Formula:

\int\limits {x}^{n} \, dx =\frac{x^{n+1}}{n+1} + C

Integrate each term:

\int\ {x^{2}} \, dx = \frac{x^{3}}{3}

\int\ {x} \, dx =\frac{x^2}{2}

\int\ {5} \, dx = 5x

\int\ {x} \, dx =\frac{x^2}{2}

\int\ {2} \, dx = 2x

Altogether this becomes:

\frac{\frac{x^3}{3} - \frac{x^2}{2}-5x }{ \frac{x^2}{2}+2x}

Your limits are 1 and 2. Substitute both numbers into the equation and minus them from each other.

x = 2:

\frac{\frac{2^3}{3} - \frac{2^2}{2}-5(2) }{ \frac{2^2}{2}+2(2)}

-

x = 1:

[tex]\frac{\frac{1}{3} - \frac{1}{2}-5 }{ \frac{1}{2}+2}[/tex]

x = 2:

\frac{\frac{8}{3} - \frac{4}{2}-10 }{ \frac{4}{2}+4}

-

x = 1: \frac{\frac{8}{3} - 2-10 }{ 2+4}

=

-1.212317928

Now convert all the answers to integers and see which one matches -1.212317928:

Because it's a negative number, we know it can only be A or B.

(A) = -1.212317928

(B) = -1.19047619

The answer is A: -\frac{3}{2} +In\frac{4}{3}

<h2>(A) - 3/2 = ln 4/3</h2>

7 0
3 years ago
Find the upper quartile of the given data below:<br> ​<br> ​ 11, 8, 5, 4, 7, 6, 9, 10, 12
JulijaS [17]

Answer:

9.5

Step-by-step explanation:

1. Arrange in ascending order

4 5 6 7 8 9 10 11 12

2. Upper Quartile formula:

3×(N+1)÷4

3. Number of terms (N) = 9

4. Apply:

3×(9+1)/4

=3×10/4

=30/4

=7.5th term

5. 7.5th term is the average of the 7th and the 8th term so

Take average

7th term is 9 and 8th term is 10

6. Find average of 9 and 10 = 9.5

7. ANSWER : THE UPPER QUARTILE IS 9.5 FOR THE GIVEN DATA

7 0
4 years ago
HELP FAST!!
Vera_Pavlovna [14]

Answer:

\frac{16}{12} =1\frac{1}{3}

1\frac{5}{8}= \frac{13}{8}

3 0
3 years ago
I need help finding the equation of the line. Could someone please help me? Thanks.
anyanavicka [17]

The slope is 2/3, and the y-intercept is 4. In slope intercept form, the equation is y=2/3x+4

3 0
3 years ago
Suppose the standard deviation of a normal population is known to be 3, and H0 asserts that the mean is equal to 12. A random sa
blagie [28]

Answer:

z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9  

p_v =P(z>1.9)=0.0287  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=12.95 represent the sample mean

\sigma=3 represent the population standard deviation for the sample  

n=36 sample size  

\mu_o =12 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:  

Null hypothesis:\mu \leq 12  

Alternative hypothesis:\mu > 12  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

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

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

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

z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9  

P-value  

Since is a right tailed test the p value would be:  

p_v =P(z>1.9)=0.0287  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.  

4 0
3 years ago
Read 2 more answers
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