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uranmaximum [27]

3 years ago

15

Pls answer WARNING : NON SENSE ANSWER REPORT TO MODERATOR

1 answer:

kotykmax [81]3 years ago

7 0

Answer:

I think its A

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Your’re in charge of evening entertainment for an important client group You use the company credit card to take their four repr

katen-ka-za [31]

Answer:

total amount paid = 32.5 + 28.9 + 24.95 = 86.35

20% of the total amount paid = 0.2 * 86.35 = 17.27

you tip 17.25$

5 0

4 years ago

Lily bought 22.17 pounds of grapefruit. The lightest grapefruit weighed 1.3 pounds. The heaviest grapefruit weighed 1.7 pounds.

nalin [4]

Answer:

<em>Approximately 15 grapefruits</em>

Step-by-step explanation:

<u>Average Value</u>

The mean or average of a sample is a number that represents the whole set by the center value of them. The formula to compute the average is

$\displaystyle \bar x=\frac{\sum x_i}{n}$

Where xi is the value of the element i and n is the number of elements in the set

We need to find an estimate of how many grapefruits Lily purchased knowing the heaviest and the lightest weight of the set. We can find the average value by using the two extreme values:

$\displaystyle \bar x=\frac{1.3+1.7}{2}=1.5$

If each grapefruit weighed 1.5 pounds, then we can estimate the 22.17 pounds of them will contain

$22.17/1.5\approx 15\ \text{grapefruits}$

3 0

3 years ago

What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?

lord [1]

the answer would be B :) hope this helps

5 0

4 years ago

Plzzzzzzzz help with your own handwritten picture<br><br> 120−96×[5 ÷ {23 − 3 × (12 − 22 − 16)}]

laiz [17]

This is what I got Hope the helped!!

Answer:

8 0

3 years ago

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n 2 if heads comes up

Artyom0805 [142]

Answer:

In the long run, ou expect to lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X $\sim$ geom(p = 1/2)

the probability function P can be computed as:

$P (X = n) = p(1-p)^{n-1}$

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

this implies that , you need to be paid $\sum \limits ^{n}_{i=1} n^2 P(X=n)$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2$ ∵ X $\sim$ geom(p = 1/2)

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =6$

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to lose $4 per game

3 0

4 years ago