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natita [175]
2 years ago
12

Voting age

Mathematics
1 answer:
Levart [38]2 years ago
4 0

Answer:

\frac{17}{64}\approx 0.27

Step-by-step explanation:

We have been given a table to voters and their ages. We are asked to find the probability that a voter is younger than 45.

Voting age         Voters

17-29                       9

30-44                      8

45-64                    32

65+                        15

We can see from our given table that age of 17 (9+8) voters is between 17 to 44 years.

To find the probability that a voter is younger than 45, we will divide 17 by total number of voters.

\text{Total voters}=9+8+32+15=64

\text{Probability that a voter is younger than 45}=\frac{17}{64}

\text{Probability that a voter is younger than 45}=0.265625

\text{Probability that a voter is younger than 45}\approx 0.27

Therefore, the probability that a voter is younger than 45 is 0.27.

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Answer:

The 99% confidence interval = (126.93,157.67)

Step-by-step explanation:

The formula for Confidence Interval =

Mean ± z × Standard deviation /√n

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Z score for 99% confidence interval = 2.56

Confidence Interval =

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Therefore, the 99% confidence interval = (126.93,157.67)

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2 years ago
How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy
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Answer:

300\; \rm lb.

Step-by-step explanation:

Let x represent the mass (in pounds) of that 15\% copper alloy required, such that the final mixture would contain 25.5\% copper by mass.

Consider: if x pounds of that 15\% copper alloy is mixed with 700 pounds that 30\% copper alloy, what would be the mass of copper in the mixture?

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Therefore, the mixture would contain (210 + 0.15\, x) \; \rm lb of copper.

The mass of that mixture would be (700 + x)\; \rm lb. The mass fraction of copper in that mixture would be:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\%.

This ratio is supposed to be equal to 25.5\%. These two pieces of equations combine to give an equation about x:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%.

\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255.

Simplify and solve for x:

210 + 0.15\, x= 0.255\, (700 + x).

(0.255 - 0.15)\, x= 210 - 0.255 \times 700.

\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300.

Therefore, 300\; \rm lb of that 15\% alloy would be required.

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