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3 years ago

PROMPT

Mathematics

1 answer:

fredd [130]3 years ago

5 0

<span>1. Photo description: A picture of the Eiffel tower, to be stuck on a mat.
Dimensions (including units): 4 in x 6 in
2. Since 2x would be added to each dimension:
Length: 6 + 2x (inches)
Width: 4 + 2x (inches)
3. Area: A = LW = (6+2x)(4+2x) square inches
4. F: (6)(4) = 24, O: (6)(2x) = 12x, I: (2x)(4) = 8x, L: (2x)(2x) = 4x^2
Polynomial expression: Adding the FOIL terms up: 4x^2 + 20x + 24
5. The area should be in square inches, since we multiplied length (in inches) by width (in inches).
6. Multiply factors using the distribution method:
(6+2x)(4+2x) = 6(4+2x) + 2x(4+2x) = 24 + 12x + 8x + 4x^2 = 24 + 20x + 4x^2
This is identical to the expression in Part 4.
7. x: 24 + 20x + 4x^2
If x = 1.0 in: Area = 24 + 20(1) + 4(1)^2 = 48 in^2
If x = 2.0 in: Area = 24 + 20(2) + 4(2)^2 = 80 in^2
8. If a white mat costs $0.03 per square inch and a black mat costs $0.05 per square inch, determine the cost of each size of black and white mat.
x Total area of mat Cost of white mat Cost of black mat
1.0 in, A = 48 in^2, (0.03)(48) = $1.44, (0.05)(48) = $2.40
2.0 in, A = 80 in^2, (0.03)(80) = $2.40, (0.05)(80) = $4.00
9. The cheapest option would be the white mat with 1-in margins on all sides, which would cost $1.44. Without any further criteria on aesthetics or size limitations, this is the most viable option.</span>

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At a new exhibit in the museum of science, people are asked to choose between 90 or 120 random draws from a machine. The machine

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Answer:

There is a probability of P₉₀=0.09853 of winning the prize with a 90-balls sample.

There is a probability of P₁₂₀=0.0853 of winning the prize with a 120-balls sample.

Step-by-step explanation:

In this problem, the balls that are removed are then replaced in the total, so the probability of obtaining a green ball is constant and is equal to:

$P_g=\frac{93}{93+65}= 0.5886$

This type of sampling should be analyzed with the binomial distribution, but given the sample size, it is convenient to use the approximation to a normal distribution.

The parameters of the normal distribution are:

$\mu=P_g*n=0.5886*90=52.9746\\\\\sigma=\sqrt{np(1-p)}=\sqrt{90*0.5886*(1-0.5886)} =4.6684$

To win the prize it is needed at least 65% of green balls in the sample, which means there have to be at least 59 green balls in the sample:

$0.65*90=58.5$

To calculate the probability of getting 59 or more green balls, we have to calculate the z-value

$z=\frac{x-\mu}{\sigma}=\frac{59-52.9746}{4.6684}= 1.29$

Then, the probability can be look up with the z-value in a normal distribution table:

$P(X \geq 59)=P(z\geq 1.29)=0.09853$

There is a probability of P=0.09853 of winning the prize with a 90-balls sample.

When the sample is of 120 balls, we have to recalculate the parameters.

Normal distribution

$\mu=P_g*n=0.5886*120=70.6320\\\\\sigma=\sqrt{np(1-p)}=\sqrt{120*0.5886*(1-0.5886)} =5.3905$

To win the prize it is needed at least 65% of green balls in the sample, which means there have to be at least 78 green balls in the sample:

$0.65*120=78$

To calculate the probability of getting 78 or more green balls, we have to calculate the z-value

$z=\frac{x-\mu}{\sigma}=\frac{78-70.6320}{5.3905}=1.37$

Then, the probability can be look up with the z-value in a normal distribution table:

$P(X \geq 78 )=P(z\geq 1.37)=0.0853$

There is a probability of P=0.0853 of winning the prize with a 120-balls sample.

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Answer:

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