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Thepotemich [5.8K]

3 years ago

5

Is the point (16,20) on line ℓ?

1 answer:

kicyunya [14]3 years ago

7 0

Answer:

YES

Step-by-step explanation:

Find the equation of the line written as, y = kx. The graph shows a proportional relationship between y and x.

Constant of proportionality/unit rate/slope (k) = rise/run = ⁵/4.

Substitute ⁵/4 in y = kx

We would have:

y = ⁵/4x.

Using the equation of the line, we can know if a given point is on the line by plugging the value of x and y coordinates of the point into the equation. If it makes the equation true, then it is a point on the line. If it doesn't make the equation true, then it isn't a point on the line.

Let's plug in (16, 20) into y = ⁵/4x.

Thus substitute x = 16 and y = 20, we have:

[tex] 20 = \frac{5}{4}(16) [/trex]

[tex] 20 = (5)(4) [/trex]

[tex] 20 = 20 [/trex]

It makes the equation true. Therefore, the point, (16, 20) is a point on line l.

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We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

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

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

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

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Sum of squares of differences = 642.1

$S.D = \sqrt{\frac{642.1}{49}} = 8.44$

We are given the following in the question:

Population mean, μ = 106

Sample mean, $\bar{x}$ = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 106\\H_A: \mu > 106$

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

Formula:

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

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$t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135$

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

Since,

$t_{stat} > t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

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