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3 years ago

Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

Mathematics

2 answers:

mixas84 [53]3 years ago

7 0

Answer by JKismyhusbandbae: 26d - 24

How to get Answer: $\mathrm{Expand}\:2\left(d-6\right):\quad 2d-12\\=2d-12-4\left(-6d+3\right)\\\mathrm{Expand}\:-4\left(-6d+3\right):\quad 24d-12\\=2d-12+24d-12\\\mathrm{Simplify}\:2d-12+24d-12:\quad 26d-24\\=26d-24$

Explanation: Use the distributive property to remove both sets of parentheses. Then combine like terms (terms involving the variable d and constant terms) to simplify the expression.

Andru [333]3 years ago

4 0

I'm assuming you want us to simplify this

remember the distributive property

a(b+c)=ab+ac

2(d-6)-4(-6d+3)

2(d-6)=2(d)+2(-6)=2d-12

-4(-6d+3)=(-4)(-6d)+(-4)(3)=24d-12

2(d-6)-4(-6d+3)=2d-12+24d-12=26d-24

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Whoevr answers this question correctly gets brainly and show your work please

blsea [12.9K]

Answer:

1/4

Step-by-step explanation:

The chance of it landing heads up once would be 1/2

The chance of it landing heads up twice would be 1/4

The chance of it landing heads up three times would be 1/8

The chance of it landing heads up four times would be 1/16

See the pattern?

Pattern: Divide the denominator by 2.

<h2>Please give brainliest!!! :) </h2>

3 0

3 years ago

The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

5 0

3 years ago

Emma made 3/4 of a pound of trail mix. If she puts 1/8 of a pound into each bag, how many bags can Emma fill?

xxMikexx [17]

3/4 divided by 1/8 would be 6 so Emma can fill 6 bags. Hope this helps and please mark brainliest!

3 0

2 years ago

Ill mark you brainliest please help !

Artemon [7]

Answer: (-2, 1)

Step-by-step explanation:

3 0

3 years ago

A bag has 3 red marbles, 2 blue, 4 greens and 1 yellow. What is the probability of pulling a blue, putting it back

Tomtit [17]

Answer: 2/25

Step-by-step explanation: Since we are replacing the marbles, these two events are independent which means that the outcome of the first does not affect the outcome of the second.

First find the probability of selecting a blue marble.

There are 2 favorable outcomes, the blue marbles,

and 10 possible outcomes, all the marbles in the bag.

So the probability of selecting a blue marble is 2/10 or 1/5.

Always reduce when finding probability

Now we find the probability of selecting a green marble.

There are 4 green marbles and 10 total so the probability of

selecting a green marble is 4/10 or 2/5.

Now we multiply the probabilities.

So we have 1/5 x 2/5 or 2/25.

3 0

3 years ago