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Andrej [43]
3 years ago
12

Herman and his dad are leaving early in the morning for his lacrosse tournament. Their house is 210 miles from the tournament. T

hey plan to stop and eat after 2 hours of driving, then compete the rest of the trip. Herman's dad plans to drive at an average speed of 60 miles per hour. How long will the second part of the trip take
Mathematics
1 answer:
Licemer1 [7]3 years ago
6 0

The second part of the trip took 1.5 hours.

Average Speed is the ratio of the total distance travelled to the total time taken to cover that distance. It is given by:

Average Speed = total distance ÷ total time

Given that the distance from the house to the tournament is 210 miles, hence:

Total distance = 210 miles

Also, The journey was in two parts, the first part took 2 hours. Let the second part take t hours. Therefore:

Total time = t + 2

The average speed of 60 miles per hour, hence:

60 = 210 ÷ (t + 2)

60(t + 2) = 210

60t + 120 = 210

60t = 90

t = 1.5 hours

Therefore the second part of the trip took 1.5 hours.

Find out more at: brainly.com/question/9834403

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Answer:

19,200 cm^2.

Step-by-step explanation:

The area = 1.2 * 1.6 = 1.92 m^2.

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List factors of 24 36 27 and 7​
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Answer:

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The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.

Step-by-step explanation: I hope this helps :).

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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
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