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4 years ago

Help me please justify each step 9x/4=-9 9x=-36 x=-4

2 answers:

6 0

$\dfrac{9x}{4}=-9\ \ \ |\text{Multiplication Property of Equality}\\\\4\cdot\dfrac{9x}{4}=4\cdot(-9)\\\\9x=-36\ \ \ \ |\text{Division Property of Equality}\\\\\dfrac{9x}{9}=\dfrac{-36}{9}\\\\x=-4$

aleksandrvk [35]4 years ago

3 0

Answer:

The justification is mentioned below.

Step-by-step explanation:

Given equation,

$\frac{9x}{4}=-9$

Multiply both sides by 4 ( multiplicative property of equality ),

$\frac{9x}{4}\times 4 = -9\times 4$

$9x=-36$

Divide both sides by 9 ( division property of equality )

$\frac{9x}{9}=\frac{-36}{9}$

$x=-4$

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3 years ago

find three consecutive integers. such that 4 times of first integer is 18 more than the sum of the second and third integer

Juliette [100K]

Three consecutive numbers are x, x+1 and x+2.

Four times the first integer is 4x

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So, we have

$4x = 2x+3+18\iff 4x=2x+21$

Subtract 2x from both sides:

$2x=21$

Divide both sides by 2:

$x=10.5$

So, you can't have three consecutive integers such that four times the first is 18 more than the sum of the other two: the three numbers would be 10.5, 11.5, 12.5.

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3 years ago

A study is performed in San Antonio to determine whether the average weekly grocery bill per five-person family in the town is s

bogdanovich [222]

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No, the sample evidence is not statistically significant (P-value = 0.125).

To reject the null hypothesis, the P-value has to be smaller than the significance level, so the significance level to reject the null hypothesis has to be 0.125 or higher.

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This is a hypothesis test for the population mean.

The claim is that the average weekly grocery bill per five-person family in San Antonio is significantly different from the national average.

Then, the null and alternative hypothesis are:

$H_0: \mu=131\\\\H_a:\mu\neq 131$

The significance level is 0.05.

The sample has a size n=50.

The sample mean is M=133.474.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.193.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.193}{\sqrt{50}}=1.583$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{133.474-131}{1.583}=\dfrac{2.474}{1.583}=1.563$

The degrees of freedom for this sample size are:

$df=n-1=50-1=49$

This test is a two-tailed test, with 49 degrees of freedom and t=1.563, so the P-value for this test is calculated as (using a t-table):

$P-value=2\cdot P(t>1.563)=0.125$

As the P-value (0.125) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the average weekly grocery bill per five-person family in San Antonio is significantly different from the national average.

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