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3 years ago

True or false? Thanks!!

Mathematics

1 answer:

statuscvo [17]3 years ago

4 0

False. 6 cannot have two y’s

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HELP I WILL GIVE BRAINLIEST!

kvv77 [185]

Answer:

1: 38

2: 56

3: 39

4: 29

5: 6

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3 years ago

Please help! Will give brainliest

jekas [21]

Answer:

5/16??? I think

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3 years ago

PLEASE HELP AND ASAP!!!! look at screenshot (10 pts)

Ne4ueva [31]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

What is an equation of the line that is parallel to y=4x−10 and passes through (1, 13) ?

klasskru [66]

If a line is parallel to another line,Then the gradients are the same.So m=4 .Using general formula for str8 line, y=mx+c
$y = 4x + c$
sub in coorfinates (1,13)
$13 = 4 + c$
So C=9.
$y = 4x + 9$

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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