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3 years ago

The box plots show the target heart rates of men 20-40 years old and men 50-70 years old.

Mathematics

1 answer:

Harman [31]3 years ago

6 0

Answer: D

Step-by-step explanation:

The range for the thing is greater

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A scientist wants to find the average mass of fish living in a large pond. So she will use the masses of a sample of fish from t

Oxana [17]

The scientist would use milligrams, grams, and kilograms.

The best procedure would be to randomly pick 52 fish from the entire pond and weigh each of them as this would be the most unbiased way to achieve the average mass

6 0

3 years ago

1. 14=7b<br> 2. -11=k/4<br> 3.6z=-6

Romashka-Z-Leto [24]

Answer:

the anwer is 14-2294-dhi2

Step-by-step explanation:

Because 14 is - 4 you multiple to 5 equal= dhi2

3 0

2 years ago

How do I solve for y?

Allisa [31]

First we know that,

3x+15=5x-65 because of vertical angle thr.

y=y because of reflex property

there are 360 degrees in a circle so..

3x+15+5x-65+2y=360

3x+15=5x-65

x=40

so each 2 of the angles are...(40*3+15)=135

Then your new equation is

135+135+2y=360

2y=90

Hope this helps :)

5 0

3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

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

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago

What is 2x=16 i to explan it

valentina_108 [34]

Answer:

2x=16

/2 /2

x=8

Step-by-step explanation:

just need to isolate the x, by dividing by two on both sides.

hope this helps :)

3 0

3 years ago

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