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3 years ago

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Why is the relationship between speed and distance linear?

1 answer:

den301095 [7]3 years ago

3 0

Answer:

Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance. For example, suppose a person were to travel 30 km/h for 4 h.

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Write the point-slope form of the line that passes through the points (-2, 1) and (0, 1). Identify (x1, y1) as (-2, 1).

kondor19780726 [428]

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ -2 &,& 1~) 
% (c,d)
&&(~ 0 &,& 1~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
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\\\\\\
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\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-1=0[x-(-2)]\implies y-1=0(x+2)$

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A car is advertised as having a gas mileage rating of at least 30 miles/gallon in highway driving. Suppose the miles per gallon

Aliun [14]

Answer:

We conclude that the car have mileage rating less than 30 miles per gallon.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30 miles/gallon

Sample:

26, 24, 20, 25, 27, 25, 28, 30, 26, 33

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{264}{10} = 26.4$

Sum of squares of differences = 110.4

$S.D = \sqrt{\frac{110.4}{9}} = 3.50$

Sample size, n = 10

Alpha, α = 0.05

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 30\text{ miles per gallon}\\H_A: \mu < 30\text{ miles per gallon}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{26.4 - 30}{\frac{3.50}{\sqrt{10}} } = -3.25$

Now,

$t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } =-1.833$

Since,

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and reject it.

We conclude that the car have mileage rating less than 30 miles per gallon.

b) We assumed that the level of significance is 0.05 that is 5%.

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3 years ago