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3 years ago

(2-7m^6)^2 i dont know this

1 answer:

7 0

(2-7m^6)^2

rearrange terms:

(2-7m^6)^2

(-7m^6 + 2)^2

expand the squares:

(-7m^6 + 2)^2

(-7m^6 + 2)(-7m^6 + 2)

distribute:

(-7m^6 + 2)(-7m^6 + 2)

-7(-7m^6 + 2) • m^6 + 2(-7m^6 + 2)

distribute:

-7(-7m^6 + 2) • m^6 + 2(-7m^6 + 2)

49m^12 - 14m^6 + 2(-7^6 +2)

distribute:

49m^12 - 14m^6 + 2(-7^6 +2)

49m^12 - 14m^6 - 14m^6 + 4

combine like terms:

49m^12 - 14m^6 - 14m^6 + 4

49m^12 - 28m^6 + 4

solution:

49m^12 - 28m^6 + 4

please mark brainliest and hope it helps

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"There are 15 questions on an exam. In how many ways can the exam be answered with exactly 8 answers correct?"

erastovalidia [21]

Answer:

The exam can be answered with exactly 8 answers correct in 6435 ways.

Step-by-step explanation:

The order is not important.

For example, answering correctly the questions 1,2,3,4,5,6,7,8 is the same outcome as answering 2,1,3,4,5,6,7,8. So we use the combinations formula to solve this problem.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

"There are 15 questions on an exam. In how many ways can the exam be answered with exactly 8 answers correct?"

Combinations of 8 questions from a set of 15. So

$C_{15,8} = \frac{15!}{8!(15-8)!} = 6435$

The exam can be answered with exactly 8 answers correct in 6435 ways.

5 0

3 years ago

PLEASEEE help with 3-4 and a-b :)

attashe74 [19]

The awnser would be-(4a+b-3) I hope this helps you

3 0

3 years ago

4.66666666667 as a fraction. (They all put 466666666667/100000000000)

natali 33 [55]

<h2>Answer:</h2><h2><em><u>Brainliest plz</u></em></h2>

Step-by-step explanation:

that is because you have to write it like this:

4.666...

in fractions its 4 and 2/3

$4 \frac{2}{3}$

7 0

3 years ago

Read 2 more answers

Nasser invested $550 at 6% simple interest for 5 years. What was his investment worth after 5

kirill [66]

Answer:

B. $715

Step-by-step explanation:

By multiplying your starting value, in this case $550, and your (simple) interest, 6%, or by 1.06 as 0.06 being your interest value and the 1.00 accommodating your starting value you will multiply, $550 × 1.06 = $583. Repeat these steps for however many years you are account for, in this case, five years. Giving you a total of $715

4 0

4 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

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 time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

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

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

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

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

3 years ago