5. (-4, 6), m = 8 8. (3,3), m=4/3

Can someone help me with this? (5+w)(w+4)

mel-nik [20]

Answer:

w^2+9w+20

Step-by-step explanation:

Apply the FOIL METHOD: (a + b) (c + d) = ac + ad + bc + bd

a = 5

b = w

c = w

d = 4

= 5w + 5 x 4 + ww + w x 4

= 5w + 5 x 4 + ww + 4w

Simplify 5w + 5 x 4 + ww + 4w:

w^2 + 9w + 20

5 0

3 years ago

Solve

polet [3.4K]

Answer:

3

4

5

9

2

Step-by-step explanation:

4 0

3 years ago

Is reading a book that is 1400 pages he will read the same number of pages each day if he reads the book in 7 days how many page

Alex777 [14]

He will read 200 pages a day because 1400/7 = 200

3 0

3 years ago

Read 2 more answers

A random sample of 21 observations is used to estimate the population mean. The sample mean and the sample standard deviation ar

stepladder [879]

Answer:

a. CI=[128.79,146.41]

b. CI=[122.81,152.39]

c. As the confidence level increases, the interval becomes wider.

Step-by-step explanation:

a. -Given the sample mean is 137.6 and the standard deviation is 20.60.

-The confidence intervals can be constructed using the formula;

$\bar X\pm \ z\frac{s}{\sqrt{n}},$

where:

$s$ is the sample standard deviation
$z$ is the s value of the desired confidence interval

we then calculate our confidence interval as:

$\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.05/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm1.960\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm8.8108\\\\\\=[128.789,146.411]$

Hence, the 95% confidence interval is between 128.79 and 146.41

b. -Given the sample mean is 137.6 and the standard deviation is 20.60.

-The confidence intervals can be constructed using the formula in a above;

$\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.01/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm3.291\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm 14.7940\\\\\\=[122.806,152.394]$

Hence, the variable's 99% confidence interval is between 122.81 and 152.39

c. -Increasing the confidence has an increasing effect on the margin of error.

-Since, the sample size is particularly small, a wider confidence interval is necessary to increase the margin of error.

-The 99% Confidence interval is the most appropriate to use in such a case.

7 0

4 years ago

At a certain elementary school, 10 percent of the fifth-grade students are members of the school band. if 12 percent of the fift

Greeley [361]

Let the 5-grade students be A in total.

Let b of these be boys, and g of these be girls, so b+g=A.

i) "At a certain elementary school, 10 percent of the fifth-grade students are members of the school band."

this means that the number of the members of the school band is :

$\frac{10}{100}A=0.1A=0.1(b+g)=0.1b+0.1g$

ii)
"12 percent of the fifth-grade boys and 8 percent of the fifth-grade girls are members of the band"

this means that the number of the band members is:
$\frac{12}{100}b+ \frac{8}{100}g=0.12b+0.08g$

equalizing these 2 ways of expressing the number of members of the band, we have:

$0.1b+0.1g=0.12b+0.08g$

$0.1g-0.08g=0.12b-0.1b$

$0.02g=0.02b$

so g=b

this means that 50% of the 5-grade students are boys.

Answer: 50%

8 0

3 years ago