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2 years ago

Help me plssssssss......

Mathematics

2 answers:

nikitadnepr [17]2 years ago

4 0

265 have a wonderful day

weqwewe [10]2 years ago

3 0

The answer is 265.

I got it by using a calculator.

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60 dollars for 12 hamburgers how much dollars per hamburger

natali 33 [55]

5. 60/12 is 5 :) hope this helps

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3 years ago

2u^2 + v when u=3 and v=7

Mariana [72]

Answer: 25

Step-by-step explanation:

There are two variables which is u and v and is given that u is 3 and v is 7 so input that into the expression and solve.

2(3)^2 + 7 Solve the exponent first

2(9) + 7 Now multiply then add

18 + 7 = 25

5 0

3 years ago

Please i need help it’s a timed test

Fofino [41]

The function can use all values of x which means x is in between -inf and +inf. The first option expresses this.

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3 years ago

A certain mathematics contest has a peculiar way of giving prizes. Five people are named as Grand Prize winners, but their finis

bixtya [17]

Answer:

The number of complete award announcements possible are 19,554,575,040.

Step-by-step explanation:

Combination is the number of ways to select k items from n distinct items when the order of selection does not matters.

Whereas permutation is the number of ways to select k item from n items when order of selection matters.

The number of people entering this year is 22.

The number of ways to select 5 people for Grand Prize is, ${22\choose 5}=\frac{22!}{5!(22-5)!} =26334$ .

The remaining number of people is, 22 - 5 = 17.

It is provided that the other 5 are selected according to an order.

The number of ways to select other 5 winners is,

$^{17}P_{5}=\frac{17!}{(17-5)!} =742560$

The total number of ways to select 10 winners of 22 is:

Total number of ways = 26334 × 742560 = 19,554,575,040.

6 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