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3 years ago

What do I need to do and what are the steps

Mathematics

1 answer:

Viktor [21]3 years ago

6 0

So allison spent 3/5 meaning she saved 2/5 because 3/5 + 2/5 = 5/5

change the 2/5 into decimals by 2 ÷ 5 = 0.4

multiply $55 with 0.4
55 x 0.4 = 22

so she ends up saving $22

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2 years ago

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2 years ago

35w+9w+35w^3+25w^3–25w^3 pls help

SVETLANKA909090 [29]

Answer:

35w^3 + 44w

Combine Like Terms:

= 35w + 9w + 35w^3 + 25w^3 - 25w^3

= (35w^3 + 25w^3 - 25w^3) +(35w + 9w)

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3 years ago

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.

4 0

3 years ago