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Vikki [24]
3 years ago
15

A 15 ounce alloy containing 30% gold was mixed with an alloy containing 5% gold to get an alloy containing 20% gold. How many ou

nces of the 5% alloy were used?
Mathematics
1 answer:
Studentka2010 [4]3 years ago
4 0

Answer:

4.5 % alloy

Step-by-step explanation:

Let x = amt of 30% alloy

The resulting total is to be 25 oz, therefore:

(25-x) = amt of 5% alloy:

A typical mixture equation:

.30x + .05(25-x) = .20(25)

.30x + 1.25 - .05x = 5 .30x - .05x = 5 - 1.25

.25x = 3.75

x = 3.75%2F.25

x = 15 oz of 30% alloy required

then

25-15 = 10 oz of 5% alloy:

Check solution

.30(15) + .05(10) = .20(25)

4.5 + .5 = 5

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The following data represent the pH of rain for a random sample of 12 rain dates in Tucker County, West Virginia. A normal proba
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Answer:

Step-by-step explanation:

n = 12

Mean = (4.58 + 5.72 + 4.77 + 4.76 + 5.19 + 5.05 + 4.80 + 4.77 + 4.75 + 5.02 + 4.74 + 4.56)/12 = 4.8925

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (4.58 - 4.8925)^2 + (5.72 - 4.8925)^2 + (4.77 - 4.8925)^2 + (4.76 - 4.8925)^2 + (5.19 - 4.8925)^2 + (5.05 - 4.8925)^2 + (4.80 - 4.8925)^2 + (4.77 - 4.8925)^2 + (4.75 - 4.8925)^2 + (5.02 - 4.8925)^2 + (4.74 - 4.8925)^2 + (4.56 - 4.8925)^2 = 1.122225

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a) Point estimate = sample mean = 4.8925

Confidence interval is written in the form,

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Margin of error = z × s/√n

Where

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 12 - 1 = 11

b) Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 2.201

Margin of error = 2.201 × 0.31/√12

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95% confidence interval = 4.8925 ± 0.197

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Lower limit = 4.8925 - 0.278 = 4.6145

We are 99% confident that the population mean of the rain water ph lies between 4.6145 and 5.1705

d) The interval gets wider as the confidence level is increased. This is logical since the test score is higher for 99% and therefore, increases the range of values. Since we want to be more confident, the range of values must be extended.

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