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Gala2k [10]
3 years ago
14

Malcolm and Theo's families are both traveling to the same vacation resort.

Mathematics
1 answer:
snow_tiger [21]3 years ago
5 0

Answer:

Malcom's family travel 15 miles per hour faster than Theo's

Step-by-step explanation:

<em>See attachment for complete question</em>

Given

Malcom's Family:

d = 65t

To determine the equation of Theo's family, we refer to the attached graph.

From the graph:

t = 1; d = 50

t = 2; d = 100

First, we determine the slope, m:

m = \frac{d_2 - d_1}{t_2 - t_1}

m = \frac{100 - 50}{2- 1}

m = \frac{50}{1}

m = 50

Next, we determine equation for Theo's family using:

d - d_1 = m(t - t_1)

d - 50 =50(t - 1)

d - 50 =50t - 50

Add 50 to both sides

d - 50 +50=50t - 50 + 50

d =50t

So, we have the following:

d = 65t --- For Malcom's family

This implies that Malcom's family travel at 65 miles per hour

d =50t --- For Theo's family

This implies that Theo's family travel at 50 miles per hour

The difference between this rates is:

Rate = 65t - 50t

Rate = 15t

<em>Which implies that Malcom's family travel 15 miles per hour faster than Theo's</em>

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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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