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vova2212 [387]
3 years ago
5

A circular running track is ¼ mile long. Elena runs on this track, completing each lap in 1/16 of an hour. What is Elena’s runni

ng speed? Include the unit of measure?
Mathematics
1 answer:
AVprozaik [17]3 years ago
3 0

Given that,

A circular running track is ¼ mile long.

Elena runs on this track, completing each lap in 1/16 of an hour

To find,

Elena’s running speed.

Solution,

Formula used :

Speed = distance/time

v=\dfrac{\dfrac{1}{4}}{\dfrac{1}{16}}\\\\=\dfrac{1}{4}\times 16\\\\=4\ mph

So, Elena’s running speed is 4 mph.

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3 and 1/3 + 4 and 5/7 =
klasskru [66]

The answer is 8 and 1/21 in fixed fraction.

and decimal form it would be 8.047619

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3 years ago
If you look at a state map with a representative fraction scale of 1:500,000; the straight-line distance between two cities is e
Leya [2.2K]

Answer:

Step-by-step explanation:

The scale factor is given as 1:500,000

i.e. original 500,000 is represented as 1 inch in the map

Hence if 12.7 inches on the map, this is a question of direct variation

So actual distance = 12.7(500000) = 6350000

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3 years ago
Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature
Vesnalui [34]

Answer:

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation  

\bar X=21.5 represent the mean for the temperatures

s=1.5 represent the sample standard deviation

n=40 sample size  

\mu_o =22 represent the value that we want to test

\alpha=0.01 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:  

Null hypothesis:\mu \geq 22  

Alternative hypothesis:\mu < 22  

If we analyze the size for the sample is > 30 and we don't know the population deviation so 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)  

t-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:  

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

P-value

We can calculate the degrees of freedom like this:

df=n-1=40-1=39

Since is a one left tailed test the p value would be:  

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0
3 years ago
Benjamin has enough icing to cover a 1/2-square-foot cake.
Serhud [2]

Answer:

A

Step-by-step explanation:

because the answer is A

5 0
3 years ago
How many solutions?​
jeyben [28]

Answer:A) no solutions

Step-by-step explanation:

1) set both problems equal to each other

2) subtract four from each side

3) subtract 2x from each side

4) 10 does not equal 0

3 0
3 years ago
Read 2 more answers
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