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3 years ago

Factor 28x² + 48x + 20

Mathematics

2 answers:

Natasha_Volkova [10]3 years ago

3 0

Answer:

4(7x+5) (x+1)

can u give me the brainiest answer

Step-by-step explanation:

Keith_Richards [23]3 years ago

3 0

Mita de 28x² es 14x

Mitad de 20 es 10

Ahora se hace 2x14x+10 La respuesta es 48x

Ahora la respuesta total es (14X+10)²

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8. Here is a graph of the equation 3x - 2y = 12.

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Answer:

A,B,D

Step-by-step explanation:

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1 year ago

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Harlamova29_29 [7]

Answer: Choice A)

y + 3 = (2/3)(x - 5)

====================================================

Explanation:

Point slope form is

y - y1 = m(x - x1)

m = 2/3 is the given slope

(x1,y1) = (5,-3) is the point the line goes through

So we have

y - y1 = m(x - x1)

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8 0

4 years ago

A magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale,

Lyrx [107]

Answer:

a) The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

b) $SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

c) The 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

Step-by-step explanation:

Notation and previous concepts

$n_1 =35$ represent the sample of ships that carry fewer than 500 passengers

$n_2 =44$ represent the sample of ships that carry 500 or more passengers

$\bar x_1 =85.82$ represent the mean sample of of ships that carry fewer than 500 passengers

$\bar x_2 =81.90$ represent the mean sample of of ships that carry 500 or more passengers

$\sigma_1 =4.55$ represent the population deviation of ships that carry fewer than 500 passengers

$\sigma_2 =3.97$ represent the sample deviation of ships that carry 500 or more passengers

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$ (1)

Part a

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

Part b: At 95% confidence, what is the margin of error?

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

The standard error is given by the following formula:

$SE=\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$

And replacing we have:

$SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

Part c: What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?

Confidence interval

Now we have everything in order to replace into formula (1):

$3.92-1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=2.009$

$3.92+1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=5.830$

So on this case the 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

7 0

3 years ago

C+12<16<br> what will be the answer

sammy [17]

Answer:

$c < 4$

Step-by-step explanation:

Move the constant to the right-hand side and change its signs:

$c < 16 - 12$

Subtract the numbers:

$c < 16 - 12 = c < 4$

3 0

3 years ago

Factor 28x² + 48x + 20​

Factor 28x² + 48x + 20