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Sloan [31]
3 years ago
15

For a science experiment, Ashley removes a cold liquid from a refrigerator and measures its temperature every 1 2 minute. Ashley

finds that the temperature increases by 1 1 4 degrees Fahrenheit (°F) between each measurement for three minutes. What is the rate per minute of the temperature increase? A) 5 8 °F per minute B) 3 4 °F per minute C) 1 3 4 °F per minute D) 2 1 2 °F per minute
Mathematics
2 answers:
Pavlova-9 [17]3 years ago
8 0

Answer:

The answer is D

Step-by-step explanation:

2

1 /2

°F per minute

1

1 /4

× 2 = 2

1 /2

elena-14-01-66 [18.8K]3 years ago
4 0
The answer will It would be D
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-0.17157287525

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3 years ago
An investigator thinks that people under the age of forty have vocabularies that are different than those of people over sixty y
Kay [80]

Answer:

t=\frac{(14-20)-0}{\sqrt{\frac{5^2}{31}+\frac{6^2}{31}}}}=-4.28  

Now we can calculate the p value with the following probability:

p_v =2*P(t_{60}  

The p value is a very low value so then we have enough evidence to reject the null hypothesis and we can conclude that people under the age of forty have vocabularies that are different than those of people over sixty years of age.

Step-by-step explanation:

Information given

\bar X_{1}=14 represent the mean for sample 1 (younger)

\bar X_{2}=20 represent the mean for sample 2 (older)  

s_{1}=5 represent the sample standard deviation for 1  

s_{f}=6 represent the sample standard deviation for 2  

n_{1}=31 sample size for the group 2  

n_{2}=31 sample size for the group 2  

t would represent the statistic

System of hypothesis

We want to test if  that people under the age of forty have vocabularies that are different than those of people over sixty years of age, the system of hypothesis are:

Null hypothesis:\mu_{1}-\mu_{2}=0  

Alternative hypothesis:\mu_{1} - \mu_{2}\neq 0  

The statistic is given by:

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=31+31-2=60  

Replacing the info given we got:

t=\frac{(14-20)-0}{\sqrt{\frac{5^2}{31}+\frac{6^2}{31}}}}=-4.28  

Now we can calculate the p value with the following probability:

p_v =2*P(t_{60}  

The p value is a very low value so then we have enough evidence to reject the null hypothesis and we can conclude that people under the age of forty have vocabularies that are different than those of people over sixty years of age.

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3 years ago
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