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2 years ago

I need help with A,B,and C

Mathematics

1 answer:

Vinil7 [7]2 years ago

8 0

Answer:

Answers!!!!!!

HOPE THIS HELPS!!!!!!!!!!

Step-by-step explanation:

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Subtract using the number line. −4−(8)

Alja [10]

The answer is -12 because the 4 and 8 are negative

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2 years ago

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A flashlight and battery sell together for $9. If the flashlight costs $5 more than the battery, what is the price of each item?

MatroZZZ [7]

The flashlight cost 7 dollars and the batter is 2 dollars

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2 years ago

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CAN YOU HELP IN MATH<br>

atroni [7]

Answer:

See below

Step-by-step explanation:

Domain: set of x values

Range: set of y values

Function: no two x values are the same

Line 1:

Domain: {3, 4, 5}

Range: {1, 2, 3, 4}

Function: no since there are two points with x-coordinate 3

Line 2:

Domain: {-1, -2, 1, 2}

Range: {1, 3}

Function: yes since all x-coordinates are different

Line 3:

Domain: {-2, -1, 0, 1, 2}

Range: {0, 1, 4}

Function: yes since all x-coordinates are different

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1 year ago

What is 45% written as a fraction

Neko [114]

There are 2 ways you can write this.

You can either do /1 or /100

$\frac{45}{1} or \frac{45}{100}$

The most used is $\frac{45}{100}$

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2 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