All categories

3 years ago

What is the quotient?

Mathematics

1 answer:

yuradex [85]3 years ago

5 0

Answer:

-1.58208955224

Step-by-step explanation:

DO THE MATH

You might be interested in

Y=x(2)-2x-3 find the value of x when y=1

Volgvan

1 = x(2) - 2x - 3
1 = 2x + -2x + - 3
1 = (2x + -2x) + (-3)
1 = -3
1 - 1 = -3 -1
0 = -4

No solutions

4 0

3 years ago

Awnser please now !!!!!!!!!!!!!!

polet [3.4K]

Answer:

(2, 0)

Step-by-step explanation:

The x intercept is when y is equal to 0.

Therefore,

8x-1/3*0=16

8x=16

x=2

The answer is (2,0)

5 0

3 years ago

Read 2 more answers

Which expression is equivalent to .........? <br> ......<br> ......<br> .....<br> .

Alchen [17]

Answer:

first one

2m=1 == 2+m-1+m

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Nima uses 2/3 cup peanuts, 1/2 cup cashews, 3/4 cups pecans, and some raisinsin a recipe that makes 2 1/4 cups of trail mix. How

Lesechka [4]

Answer:

8/27 cups

Step-by-step explanation:

Nima uses 2/3 cups of peanuts to make a recipe of 2 1/4 = 9/4 cups of trail mix. If we want to find how many cups of peanuts are there per cup of trail mix, we can do a rule of three:

If 9/4 cups of trail mix need 2/3 cups of peanuts, 1 cup of trail mix need X cups of peanuts:

9/4 cups of trail mix -> 2/3 cups of peanuts

1 cup of trail mix -> X cups of peanuts

9/4 / 1 = 2/3 / X

9/4 = 2/(3*X)

3*X*9 = 4*2

27*X = 8

X = 8/27 cups

7 0

3 years ago

The Resource Conservation and Recovery Act mandates the tracking and disposal of hazardous waste produced at U.S. facilities. Pr

aniked [119]

Answer:

(a) $E(X) = 0.383$

The expected number in the sample that treats hazardous waste on-site is 0.383.

(b) $P(x = 4) = 0.000169$

There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Step-by-step explanation:

Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.

N = 209

Only eight of these facilities treated hazardous waste on-site.

r = 8

a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site?

n = 10

The expected number in the sample that treats hazardous waste on-site is given by

$E(X) = \frac{n \times r}{N}$

$E(X) = \frac{10 \times 8}{209}$

$E(X) = \frac{80}{209}$

$E(X) = 0.383$

Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.

b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site

The probability is given by

$P(x = 4) = \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}}$

For the given case, we have

N = 209

n = 10

r = 8

x = 4

$P(x = 4) = \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}}$

$P(x = 4) = \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}}$

$$ P(x = 4) = \frac{70 \times 84944276340}{35216131179263320}$

$P(x = 4) = 0.000169$

Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

8 0

3 years ago