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2 years ago

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Consider the rates 45 mile in 8 minutes and 4 minutes to travel 25 mile. Select the statement(s) that are true. Multiple select

question. Cross out A) The rate 45 mile in 8 minutes has a unit rate of 10 miles per minute. Cross out B) The rate 45 mile in 8 minutes has a unit rate of 110 mile per minute. Cross out C) The rate 4 minutes to travel 25 mile has a unit rate of 10 minutes per mile. Cross out D) The rate 4 minutes to travel 25 mile has a unit rate of 110 minute per mile. Cross out E) Both rates are equivalent. Cross out F) The rates are not equivalent.

1 answer:

icang [17]2 years ago

7 0

Answer:

F) The rates are not equivalent.

Step-by-step explanation:

Consider the rates 45 mile in 8 minutes and 4 minutes to travel 25 mile. Select the statement(s) that are true. Multiple select question.

We find the unit rates

a) 45 mile in 8 minutes , we find the unit rate in miles per minute

8 minutes = 45 miles

1 minute = x

8 × x = 45 miles × 1 minute

x = 45 miles/8 minutes

x = 5.625 miles/ minute

b) 4 minutes to travel 25 miles. Find the unit rate is minutes per mile

25 miles = 4 minutes

1 mile = x

25miles × x = 4 minutes × 1

x = 4 minutes/25 miles

x = 0.16 minutes/mile

Hence, the correct option is

F) The rates are not equivalent.

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Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

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