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4 years ago

I need help. with proof/work

Mathematics

1 answer:

tankabanditka [31]4 years ago

4 0

So, we know these three things:

$1. \text{ } a^b*a^c=a^{b+c} \\
2. \text{ } (a^b)^c=a^{bc}$

So, we have:

$12^{11}*(12^2)^4=\\
12^{11}*12^{8} =\\
12^{11+8}=\\
12^{19}$

You might be interested in

Evaluate b-(-1/8)+c where b=2 and c=-7/4

tatiyna

Answer:

3/8

Step-by-step explanation:

Evaluate b - -1/8 + c where b = 2 and c = -7/4:

b - (-1)/8 + c = 2 - (-1)/8 - 7/4

Put 2 + 1/8 - 7/4 over the common denominator 8. 2 + 1/8 - 7/4 = (8×2)/8 + 1/8 + (2 (-7))/8:

(8×2)/8 + 1/8 - (7×2)/8

8×2 = 16:

16/8 + 1/8 - (7×2)/8

2 (-7) = -14:

16/8 + 1/8 + (-14)/8

16/8 + 1/8 - 14/8 = (16 + 1 - 14)/8:

(16 + 1 - 14)/8

16 + 1 = 17:

(17 - 14)/8

| 1 | 7

- | 1 | 4

| 0 | 3:

Answer: 3/8

8 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

What is the tangent ratio for ∠E

Stells [14]

Tan E = 6/8 = 3/4

answer
3/4

3 0

3 years ago

Read 2 more answers

What are the values of x and y in the figure shown below?

n200080 [17]

First let's find y. We know the sum of angles equals 180, so if we add the angles of ABC, we'd get 180.
30+y+80+y=180
110+2y=180
2y=70
y=35

You could solve x ,but the only response with y=35 is c. But to solve for x, you'd ad 35 35 and xwhich would equal 180.
35+35+x=180
70+x=180
x=110

5 0

3 years ago

The longest side of a right triangle is 25. If one of the other sides is 5, which measure is the length of the missing side?

jasenka [17]

Pythagorean Theorem A2+B2=C2

625=25+C2

600=C2

Square root of 2 each side

10 Square root of 6 Or 24.494897

6 0

3 years ago