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loris [4]
3 years ago
8

Sasha has 3.20 in U.S. coins. She has the same number of quarters and nickels. What is the greatest number of quarters she could

have?
Mathematics
2 answers:
RSB [31]3 years ago
8 0

Answer:

10 quarters = $2.50

10 nickels = $0.50

that leaves $0.20 for other coins (dimes / pennies)

Step-by-step explanation:

First, suppose she has only quarters and nickels and no other coins.  Then if C is the identical number of coins of each type, then 5C + 25C = 320, so 30C = 320 and 3C = 32, but there is no integer solution to this.  So she must have at least one other type of coin.

Assume she has only quarters, nickels, and dimes.  Then if D is the number of dimes, 5C +  25C + 10D = 320, which means 30C + 10D = 320, or 3C + D = 32.  The smallest D can be is 2, leaving 3C = 30 and thus C = 10.  So in this scenario she would have 10 quarters, 10 nickels, and two dimes to make $2.50 + $0.50 + $0.20 = $3.20.

This has to be the highest number, because if she had 11 quarters and 11 nickels, that alone would add up to 11(0.25) + 11(0.05) = $3.30, which would already be too much.

Nataly_w [17]3 years ago
3 0

Answer:

2.50 coins

Step-by-step explanation:

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3 years ago
What is the answer??
Alchen [17]

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4 0
3 years ago
A random sample of 400 voters in a certain city are asked if they favor an additional 4% gasoline tax to provide badly needed re
Norma-Jean [14]

Answer:

A) α = 0.04136

B) β = 0.00256

Step-by-step explanation:

We are given;

Sample size; n = 400

Proportion; p = 60% = 0.6

Formula for mean is;

μ = np

μ = 400 × 0.6 = 240

Standard deviation is given by;

σ = √npq

Where q = 1 - p = 1 - 0.6 = 0.4

σ = √(400 × 0.6 × 0.4)

σ = √96

σ = 9.8

A) our null hypothesis is at p = 0.6

Probability of making a type I error means we reject the null hypothesis when it is true.

This can be expressed in reference to the question as;

α = P(x < 220) + P(x > 260) all at p = 0.6

Now,

P(x < 220) = z = (x¯ - μ)/σ = (220 - 240)/9.8 = -2.04

Also;

P(x > 260) = z = (260 - 240)/9.8 = 2.04

Now, from z-distribution table probability of a z-score of -2.04 is 0.02068.

Also, probability of z-score of 2.04 is (1 - P(z < 2.04) = 1 - 0.97932 = 0.02068

Thus;

α = 0.02068 + 0.02068

α = 0.04136

B) Type II error occurs when we fail to reject the null hypothesis even though it's false.

In this case our alternative hypothesis is at p = 48% = 0.48

Thus;

μ = np

μ = 400 × 0.48 = 192

Standard deviation is given by;

σ = √npq

Where q = 1 - p = 1 - 0.48 = 0.52

σ = √(400 × 0.48 × 0.52)

σ = √99.84

σ = 9.992

Type II error would be given by;

β = [((x1¯ - μ)/σ) < z > ((x2¯ - μ)/σ)]

β = [((220 - 192)/9.992) < z > ((260 - 192)/9.992)]

β = (2.8 < z > 6.81)

Rearranging this gives us;

β = P(z < 6.81) - P(z < 2.8)

From z-distribution tables, we have;

β = 1 - 0.99744

β = 0.00256

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