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Sindrei [870]
3 years ago
10

Write an equation for the line, in point-slope form, that passes through the following point and has the following slope:slope:-

4 point: (1, -2)
Mathematics
1 answer:
enyata [817]3 years ago
5 0

Answer:

y+2=-4(x-1)

Step-by-step explanation:

y-y1=slope(x-x1)

just plug in the numbers you were given and youre good to go

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How do you get 6,300,00 using the digits 1-7 to creat a seven-digit number that can be rounded to 6,300,000?
Tomtit [17]
In the problem, how do you get 6,300,000 using the digits 1-7 to create a seven-digit number that can be rounded to 6,300,000. We can use the numbers according to the rule of rounding values, 0-4 and 5-9. Hence the numbers are:
1. 6, 290,000 = 6, 300, 000
2. 6, 310, 000 = 6, 300, 000 Therefore the numbers concerned in the given value is in the place order of ten thousand which will determine the hundred thousands’ value.  



6 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
3 years ago
Why should people conserve energy? (Please don’t get the answer form the internet)
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6. There are four-wheel trucks and six-
Nat2105 [25]

Answer:

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Step-by-step explanation:

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W is greater than -5 and less than 0
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Yessndndndd dndndndndndndndndndd
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