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3 years ago

Which expression is equivalent to (x+1)^2-9/x+4

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

6 0

The answer should be 3x^-7/4x

enot [183]3 years ago

4 0

Answer:

(x-2)

Step-by-step explanation:

Given expression is $\frac{(x+1)^2-9}{x+4}$

Now we factor the given expression

LEts factor the numerator

(x+1)^2 -9

9 can be written as 3^2

(x+1)^2 - 3^2

Now we apply a^2 - b^2 formula

a^2 - b^2 = (a+b)(a-b)

(x+1)^2 - 3^2 = (x+1+3)(x+1-3)=(x+4)(x-2)

now we replace it in our problem

$\frac{(x+1)^2-9}{x+4}=\frac{(x+4)(x-2)}{x+4}$

Cancel out x+4 from top and bottom

So answer is (x-2)

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3 years ago

5. The length of a rectangular swimming pool is 50 yards, and the width of the pool is 16

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3 years ago

An educator claims that the average salary of substitute teachers in school districts is less than $60 per day. A random sample

valentina_108 [34]

Answer:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Step-by-step explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Replacing we got:

$\bar X=58.875$ represent the mean

$s=5.083$ represent the sample standard deviation for the sample

$n=8$ sample size

$\mu_o =60$ represent the value that we want to test

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:

Null hypothesis: $\mu \geq 60$

Alternative hypothesis: $\mu < 60$

The statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info we got:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

3 0

3 years ago

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alex41 [277]

Answer:

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3 years ago