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valentina_108 [34]

3 years ago

9

Michael likes to go hiking. He can climb 400 ft. every 2 hours. Which equation models the relationship between the time Michael

hikes, x, and the distance he climbs, y?

1 answer:

crimeas [40]3 years ago

5 0

<u>Answer:</u>

Below!

<u>Step-by-step explanation:</u>

The word "equation" defines that an expression must have an <em><u>equal sign.</u></em><em> </em>This means that the equation is <em><u>y = x.</u></em> The variable 'y' represents <em><u>the distance Michael has climbed.</u></em> The variable 'x' represents <em><u>the time mike hiked.</u></em>

=> y = x
=> 400 ft = 2 hours
=> 200 ft = 2/2 hours
=> <u>200 ft = 1 hour</u>

<u>Hence, 200 ft = 1 hour models the relationship between the time Michael hikes, x, and the distance he climbs, y.</u>

Hoped this helped.

$BrainiacUser1357$

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The conclusion

There is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

Step-by-step explanation:

From the question we are told that

The data is

Car Ages 4 0 8 11 14 3 4 4 3 5 8 3 3 7 4 6 6 1 8 2 15 11 4 1 6 1 8

Taxi Ages 8 8 0 3 8 4 3 3 6 11 7 7 6 9 5 10 8 4 3 4

The level of significance $\alpha = 0.05$

Generally the null hypothesis is $H_o : \mu_1 - \mu_2 = 0$

the alternative hypothesis is $H_a : \mu_1 - \mu_2 > 0$

Generally the sample mean for the age of cars is mathematically represented as

$\= x_1 = \frac{\sum x_i }{n}$

=> $\= x_1 = \frac{4+ 0+ 8 +11 + \cdots + 8
}{27}$

=> $\= x_1 = 5.56$

Generally the standard deviation of age of cars

$\sigma _1 = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }$

=> $\sigma _1 = \sqrt{\frac{(4 - 5.56)^2 + (0 - 5.56)^2+ (8 - 5.56)^2 + \cdots + 8}{ 27} }$

=> $\sigma _1 = 3.88$

Generally the sample mean for the age of taxi is mathematically represented as

$\= x_2 = \frac{\sum x_i }{n}$

=> $\= x_2 = \frac{8 +8 +0 + \cdots + 4
}{20}$

=> $\= x_2 = 5.85$

Generally the standard deviation of age of taxi

$\sigma _2 = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }$

=> $\sigma _2 = \sqrt{\frac{(8 - 5.85)^2 + (8 - 5.85)^2+ (0 - 5.85)^2 + \cdots + 8}{ 20} }$

=> $\sigma _2 = 2.83$

Generally the test statistics is mathematically represented as

$t = \frac{(\= x_ 1 - \= x_2 ) - 0}{\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2} } }$

=> $t = \frac{(5.56 - 5.85 ) - 0}{\sqrt{\frac{(3.88)^2}{27} + \frac{(2.83)}{20} }}$

=> $t = -0.30$

Generally the degree of freedom is mathematically represented as

$df = n_1 + n_2 -2$

$df = 27 + 20 -2$

$df = 45$

From the t distribution table the $P(t > t )$ at the obtained degree of freedom = 45 is

$P(t > -0.30 ) = 0.61722067$

So the p-value is

$p-value = P(t > T) = 0.61722067$

From the obtained values we see that the p-value > $\alpha$ hence we fail to reject the null hypothesis

Hence the there is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

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