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3 years ago

WHEN PARKING NEAR A CORNER YOU MAY PARK YOUR VEHICLE NO CLOSER THAN HOW MAY FEET?

Mathematics

2 answers:

natta225 [31]3 years ago

3 0

You may park 20 feet from a crosswalk.

Temka [501]3 years ago

3 0

Where I live, you must be at least 19 feet from a crosswalk. It all depends on where you live, because the laws and rules differ.

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Given the lengths of two sides of a triangle, find the range for the length of the third side (between what two numbers should t

kobusy [5.1K]

The shortest the third side can be is the difference of the other two sides:
.. 23.6 -11.5 = 12.1
The longest the third side can be is the sum of the other two sides:
.. 23.6 +11.5 = 35.1

12.1 < third side < 35.1

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3 years ago

How to find the area of a polygon

Arturiano [62]

To find the area of a polygon use the formula Area = 1/2 x perimeter x apothem

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3 years ago

Find the discount on a 15$ Tshirt its on sale for 10% off

Elina [12.6K]

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Step-by-step explanation:

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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

Divide 180 degrees by the ratio 4:5:9, Help Please

mixas84 [53]

Add all the numbers in the ratio together to get 18 then divide 180 by 18 to equal 10. then times that by each individual number in the ratio. so the answer is 40:50:90

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3 years ago