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vivado [14]
3 years ago
15

Astronauts on the International Space Station commonly face a health issue that some people on Earth experience as well. Researc

h conducted on this health issue has helped solve problems on Earth.
What is this health issue?
(1 point)
a weightlessness

b hearing and sight loss

c loss of bone and muscle density

d obesity
Mathematics
2 answers:
elena-14-01-66 [18.8K]3 years ago
8 0
C is the correct answer
Ludmilka [50]3 years ago
4 0

Answer:

C

Step-by-step explanation:

In space you have to exercise 24/7, because if you don't your bones will decay little by little, the same goes for earth, it is a condition that happens if you don't move or exercise.

If possible; can I get brainliest, I mean if it's correct

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North America burned 4.5 x 1016 btu of petroleum in 1998. At this rate, how many BTU's will be burned in 9 year?
nikklg [1K]
Hello. Sonny solving this equation, multiplying 4.5 by 1016 will give us 4,572 BTU's a year. However, we are looking for the amount looking for the amount that will burn in 9 years. When you multiply 4,572 by 9, you get 41,148. 41,148 BTU's will be burned in 9 years.
4 0
3 years ago
We're testing the hypothesis that the average boy walks at 18 months of age (H0: p = 18). We assume that the ages at which boys
marusya05 [52]

Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation    

\bar X=19.2 represent the battery life sample mean    

\sigma=2.5 represent the population standard deviation    

n=25 sample size    

\mu_o =18 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)    

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

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:    

Null hypothesis:\mu = 18    

Alternative hypothesis:\mu \neq 18    

And for part III we have a one tailed test with the following hypothesis:

Null hypothesis:\mu \leq 18    

Alternative hypothesis:\mu > 18  

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".    

3) Calculate the statistic    

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

z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4    

4) P-value    

First we need to calculate the degrees of freedom given by:  

df=n-1=25-1=24  

Since is a two tailed test for parts I and II, the p value would be:    

p_v =2*P(t_{(24)}>2.4)=0.0245

And for part III since we have a one right tailed test the p value is:

p_v =P(t_{(24)}>2.4)=0.0122

5) Conclusion    

I. This finding is significant for a two-tailed test at .05.

Since the p_v. We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the p_v >\alpha. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the p_v >\alpha. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only

6 0
3 years ago
A store sells
user100 [1]
C
Hope this helps!!!!!
7 0
3 years ago
What’s the correct answer
Trava [24]
I think it’s B but I’m not sure
6 0
3 years ago
If mark can make 42 cakes in 7 days how many can he make in 5 days
ipn [44]
42%7=6. 6 times 5 is 30. 30 cakes in 5 days.
6 0
3 years ago
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