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andreev551 [17]
3 years ago
15

Bob and sarah need to paint a fence. bob can paint the fence in 4 hours and sarah can paint the fence in 2.5 hours. if they work

together, how long will it take them to paint the fence?
Mathematics
1 answer:
zysi [14]3 years ago
7 0
If Bob can paint the fence in 4 hours, then in 1 hour he gets 1/4 of the fence painted.  If Sarah can paint the fence in 2.5 hours, then she gets 2/5 of the fence painted in one hour (1/2.5 = 2/5).  We add those amounts together and set them equal to 1/x, x being the the total time it will take them to paint the whole fence.  1/4 + 2/5 = 1/x.  Find a common denominator first, then add the left side together.  5/20 + 8/20 = 1/x.  13/20 = 1/x.  Cross-multiply to get 13x = 20 and x = 1.54 hours, just a bit over an hour and a half.
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2) A gym charges a $30 membership fee and $10 per month. Write an
kari74 [83]

C = 10m + 30, where C is total cost and m is # of months

C = 10(5) + 30

C = 50 + 30

C = $80

$160 = 10m + 30

160 - 30 = 10m

130/10 = m

m = 13 months

Hope that helps

5 0
2 years ago
A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears
marta [7]

Answer:

(a) The probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b) <em>n</em> = 28.09

Step-by-step explanation:

The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.

It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of <em>Y</em>ₙ are:

\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}

(a)

For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:

\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\

Compute the probability that <em>Y</em>₂₀ exceeds 1000 as follows:

P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z>  4.47)\\=1-P(Z

**Use a <em>z </em>table for probability.

Thus, the probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b)

It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.

P(Y_{n}>1000)=0.99\\1-P(Y_{n}

The value of <em>z</em> for which P (Z < z) = 0.01 is 2.33.

Compute the value of <em>n</em> as follows:

z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n}  \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0

The last equation is a quadratic equation.

The roots of a quadratic equation are:

n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of <em>n</em> = 28.09.

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3 years ago
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Answer:

at least 25

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Answer:

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Step-by-step explanation:

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