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3 years ago

Use long division to find the quotient below.

Mathematics

2 answers:

Pachacha [2.7K]3 years ago

8 0

Answer:

Number B: 4x+1

Step-by-step explanation:

I used synthetic division instead, But hope this helps.

Brut [27]3 years ago

5 0

(4x² - 7x - 2) ÷ (x-2)

I have placed the terms corresponding to each variable, for x squared I put 4, for x - 7 and for c (constant) I put - 2.

We go down 4, which will multiply to 2.

¿Why 2?

This is obtained by matching the denominator to 0.

$x-2=0\rightarrow \boxed{x=2}$

Then we multiply 4 by 2, give us 8, add this value to the variable x (- 7), we have 1.

Multiply 1 by 2, give us 2, add this value to the constant (- 2), we have 0. ¡Finish!

Answer: 4x + 1

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Step-by-step explanation:

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3 years ago

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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

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3 years ago