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3 years ago

What is the rate of change of y with respect to x for this function?

Mathematics

2 answers:

loris [4]3 years ago

8 0

bearing in mind that the average rate of change = slope, and for that we can simply use two points off the table.

$\bf \begin{array}{|cc|ll} \cline{1-2} x&y\\ \cline{1-2} -2&12\\ 0&3\\ 3&-10.5\\ 7&-28.5\\ \cline{1-2} \end{array}\qquad \qquad \begin{array}{llll} (\stackrel{x_1}{-2}~,~\stackrel{y_1}{12})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{-10.5}) \\\\\\ slope \implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-10.5-12}{3-(-2)} \\\\\\ \cfrac{-10.5-12}{3+2}\implies \cfrac{-22.5}{5}\implies -\cfrac{4.5}{1}\implies -4.5 \end{array}$

lisabon 2012 [21]3 years ago

7 0

Answer:

-4.5

Step-by-step explanation:

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5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g

anygoal [31]

Answer:

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.

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

$Mean =\displaystyle\frac{1117}{10} = 111.7$

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:

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

Putting all the values, we have

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

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 accept the alternate hypothesis.

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

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2 years ago

4. What's the least common multiple of each group of numbers?

mina [271]

Answer:

D. 2,5, and 6

Step-by-step explanation:

Because if u mutiply 2*5 = 10 and 2*6 is 12 as 5*6 is 30 and 2*9 is 18 .

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3 years ago

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Vladimir79 [104]

Answer:

Step-by-step explanation:

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Alenkasestr [34]

Answer:

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Step-by-step explanation:

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2 years ago

I need answer Immediately pls!!!!!!

Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

$n(U)=27$

$n(A)=7$

$n(B)=18$

$n(A'\cap B')=7$

We know that,

$n(A\cup B)=n(U)-n(A'\cap B')$

$n(A\cup B)=27-7$

$n(A\cup B)=20$

Now,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=7+18-n(A\cap B)$

$n(A\cap B)=7+18-20$

$n(A\cap B)=25-20$

$n(A\cap B)=5$

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

$\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}$

$\text{Probability}=\dfrac{5}{27}$

Therefore, the required probability is $\dfrac{5}{27}$ .

3 0

2 years ago