Find the slope of the line -x = -3y + 9

Find the slope giving brainlest

NNADVOKAT [17]

Answer:

The slope of problem 11 is

$- \frac{2}{5}$

The slope of problem 10 is

$4$

3 0

2 years ago

First one right gets BRAINLIEST!!<br><br> If f(x)=x²+2x-3 and g(x)=x²-9, find (f/g) 4 and (f+g)(4)

Rasek [7]

Answer:

3 and 28

Step-by-step explanation:

note ( $\frac{f}{g}$ )(x ) = $\frac{f(x)}{g(x)}$

$\frac{f(x)}{g(x)}$

= $\frac{x^2+2x-3}{x^2-9}$ and

( $\frac{f}{g}$ )(4) = $\frac{4^2+2(4)-3}{4^2-9}$ = $\frac{16+8-3}{16-9}$ = $\frac{21}{7}$ = 3

-------------------------------------------------------

(f + g)(x) = f(x) + g(x)

f(x) + g(x)

= x² + 2x - 3 + x² - 9 = 2x² + 2x - 12 and

(f + g)(4) = 2(4)² + 2(4) - 12 = 2(16) + 8 - 12 = 32 + 8 - 12 = 28

6 0

3 years ago

(PLEASE HELP) To feed 7 horses for 4 days, Mark needs 420 kg of grass. How many kilograms of grass does Mark need to feed 9 hors

DerKrebs [107]

Answer:

810kg is the answer

Step-by-step explanation:

Hope this helps:)

7 0

2 years ago

Read 2 more answers

Please please help me

Nataly_w [17]

Answer:

The average rate of change is 4.

Step-by-step explanation:

It is 4 because when you do the formula to find the rate of change it comes out to be 4.

8 0

3 years ago

A sanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. In a random sam

soldi70 [24.7K]

Answer:

a) $52.8-2.02\frac{3.9}{\sqrt{40}}=51.554$

b) $52.8+2.02\frac{3.9}{\sqrt{40}}=54.046$

c) $52.8-2.71\frac{3.9}{\sqrt{40}}=51.129$

d) $52.8+2.71\frac{3.9}{\sqrt{40}}=54.471$

e) Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X =52.8$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=3.9 represent the sample standard deviation

n=40 represent the sample size

a) What is the lower limit of the 95% interval? Give your answer to three decimal places

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=40-1=39$

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: "=-T.INV(0.025,39)".And we see that $t_{\alpha/2}=2.02$

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

$52.8-2.02\frac{3.9}{\sqrt{40}}=51.554$

b) What is the upper limit of the 95% interval? Give your answer to three decimal places

$52.8+2.02\frac{3.9}{\sqrt{40}}=54.046$

So on this case the 95% confidence interval would be given by (51.554;54.046)

c) What is the lower limit of the 99% interval? Give your answer to three decimal places

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,39)".And we see that $t_{\alpha/2}=2.71$

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

$52.8-2.71\frac{3.9}{\sqrt{40}}=51.129$

d) What is the upper limit of the 99% interval? Give your answer to three decimal places

$52.8+2.71\frac{3.9}{\sqrt{40}}=54.471$

So on this case the 99% confidence interval would be given by (51.129;54.471)

e) Consider the claim that the mean amount of garbage per bin is 54.1985 pounds. Is the following statement true or false? The decision about the claim would depend on whether we use a 95% or 99% confidence interval: True/False

Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

3 0

4 years ago