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goldfiish [28.3K]
3 years ago
9

An airplane flew 1113 miles from San Antonio Texas to Saint Paul Minnesota the trip to Q point 75 hours which is closest to the

average speed in miles per hour of the airplane
Mathematics
2 answers:
ycow [4]3 years ago
7 0

Answer:

14.84miles/hour

Step-by-step explanation:

Average speed is given by

Average speed= total distance/time taken

Then, the total distance traveled is 1113miles

And the total time taken is 75hours

Then

A.S=d/t

A.S=1113/75

A.S=14.84miles/hour

statuscvo [17]3 years ago
5 0

Answer:

14.84miles/hour

Step-by-step explanation:

Average speed is defined as the rate of change in distance of a body.

Average speed = Distance/Time taken

Since the airplane flew 1113 miles from San Antonio Texas to Saint Paul Minnesota;

Distance covered = 1113miles

Time taken =75hours

Substituting the formulas into the formula for calculating speed, we have;

Speed = 1113miles/75hours

Speed = 14.84miles/hour

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Answer:

a-1) Reject H0 if zcalc > 1.645

a-2) z=\frac{0.316 -0.28}{\sqrt{\frac{0.28(1-0.28)}{114}}}=0.8561  

a-3) ii. False

b) ii. No

c) iii. n π > 10 and n(1 − π ) > 10

Step-by-step explanation:

1) Data given and notation

n=114 represent the random sample taken

X=36 represent the people that were issued to a single-earner family or individual

\hat p=\frac{36}{114}=0.316 estimated proportion of people that were issued to a single-earner family or individual

p_o=0.28 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

(a-1) H0: π ≤ .28 versus H1: π > .28. Choose the right option. Reject H0 if zcalc > 1.645 Reject H0 if zcalc < 1.645 a b

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.28.:  

Null hypothesis:p\geq 0.28  

Alternative hypothesis:p < 0.28  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

The rejection zone would be on this case :

Reject H0 if zcalc > 1.645

Since is a right tailed test

(a-2) Calculate the test statistic. (Round intermediate calculations to 2 decimal places. Round your answer to 4 decimal places.) Test statistic

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.316 -0.28}{\sqrt{\frac{0.28(1-0.28)}{114}}}=0.8561  

(a-3) The null hypothesis should be rejected.

False, since our calculated value is less than our critical value we Fails to reject the null hypothesis

(b) Is this a close decision?

False the calculated value is significantly less than the critical value so we FAIL to reject the null hypothesis with enough confidence.

(c) State any assumptions that are required.

In order to satisfy the conditions we need the following two requirements:

iii. n π > 10 and n(1 − π ) > 10

And are satisfied:

114*0.28=31.92>10

114(1-0.28)=82,08>10

7 0
3 years ago
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