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Charra [1.4K]
3 years ago
9

-6-7(c+10) A. 64-7c B. -76-7C C. 4-13C D. -16-13c Help?

Mathematics
1 answer:
Maurinko [17]3 years ago
6 0

Answer:

B. -76-7C

Step-by-step explanation:

-6-7(c+10)

Distribute the 7

-6-7c-70

Combine like terms

-7c-76

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erma4kov [3.2K]
The perfect square is answer A

8 0
3 years ago
What is the total amount that First Consumer Bank will receive after lending Jane $7,000 for three years at an interest rate of
8_murik_8 [283]
Given:
principal = 7,000
interest rate = 5% compounded annually
term = 3 years

A = P (1 + r/n)^nt

A = future amount to be received by First Consumer Bank
P = loan principal
r = rate
n = number of times compounded in a year
t = term

A = 7,000 ( 1 + 5%/1)^1x3
A = 7,000 (1.05)³
A = 7,000 (1.157625)
A = 8,103.375

First Consumer Bank will receive 8,103.375 from Jane after lending 7,000 for 3 years compounded annually at 5%.
6 0
3 years ago
A survey of 120 students found 100 like music, 80 like dancing, and 75 like both. How many like music or dancing (or both)?
Fed [463]

Answer:

75 students out of 120

Step-by-step explanation:

75/120

4 0
3 years ago
If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calcul
Wittaler [7]

Answer:

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

3 failures every twenty weeks

This means that for 1 week, \mu = \frac{3}{20} = 0.15

Calculate the probability that there will not be more than one failure during a particular week.

Probability of at most one failure, so:

P(X \leq 1) = P(X = 0) + P(X = 1)

Then

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.15}*0.15^{0}}{(0)!} = 0.8607

P(X = 1) = \frac{e^{-0.15}*0.15^{1}}{(1)!} = 0.1291

Then

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.8607 + 0.1291 = 0.9898

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

6 0
3 years ago
HELP PLZ IM FAILING MATH TY :'D <br> PLEASE STOP COMMENTING FAKE ANSWERS THIS IS MY REAL SCHOOL WORK
Rzqust [24]

Answer:

5cm

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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