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galben [10]
3 years ago
8

Manny makes $8.75 an hour working at the video store. His paycheck shows that he worked 37.5 hours over the past week. How much

money did Manny make? (Not rounded to the nearest cent)
Mathematics
1 answer:
aleksley [76]3 years ago
6 0
Y=8.75x
So (8.75)(37.5)= 328.125$. All you have to do is multiply his hours by his pay.
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Answer: 2x+5y=100 and x+y=38

Step-by-step explanation:

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Just need help with number 1. Thanks!
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For 1 you can use 2 or -4, either one works.

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suppose you have a dime, two pennies, and a quarter. one of the pennies was minted in 1976, and the other one was minted in 1992
notsponge [240]

The reason that the answers in part (a) and part (b) are not the same is because Q1 and Q2 are treated as two different coins in part (a) whereas they are effectively treated as the same coin in part (b). The reason they are treated as one coin is because they both contribute the same amount to the sum of money.

<h3>How to Solve Counting Problems?</h3>

A) If you choose at least one coin, this means you could choose 1, 2, 3 or 4 coins. Let us label the dime as D, the penny as P, the quarter minted in 1976 as Q1 and the quarter minted in 1992 as Q2.

Now, if you choose one coin, you could choose either D, P, Q1, or Q2. This gives us 4 possible sets.

If you choose two coins, you choose the following sets of coins: DP, DQ1, DQ2, PQ1, PQ2, Q1Q2. This gives us 6 possible sets.

If you choose three coins, you could the following sets of coins: DPQ1, DPQ2, DQ1Q2, PQ1Q2. This gives us 4 possible sets.

If you choose four coins, you can only choose DPQ1Q2. This gives us 1 possible set.

Therefore, the total number of different sets of coins you can form is 4 + 6 + 4 + 1 = 15 different sets of coins can be formed.

b) If you choose at least one coin, this means you could choose 1, 2, 3 or 4 coins.

If you choose one coin you could choose either D, P, Q1, or Q2. However, since Q1 and Q2 give us the same sum, they are effectively the same set. This gives us 3 possible sums (ten cents, one cent, or twenty-five cents.)

If you choose two coins, you could choose the following sets of coins (since Q1 and Q2 are the same coin value, we can say Q for any instance where either one of these coins would be a possibility): DP, DQ, PQ, Q1Q2. This gives us 4 possible sums (11 cents, 35 cents, 26 cents, or fifty cents.)

If you choose three coins, you could choose the following sets of coins (since Q1 and Q2 are the same coin value, we can say Q for any instance where either one of these coins would be a possibility): DPQ, DQ1Q2, PQ1Q2. This gives us 3 possible sums (36 cents, 60 cents, or 51 cents).

If you choose four coins, you can only choose DPQ1Q2. This gives us 1 possible sum (61 cents.)

Therefore, the total number of different sums of coins you can form is 3 + 4 + 3 + 1 = 11 different sums of money can be produced.

c) The reason that the answers in part (a) and part (b) are not the same is because Q1 and Q2 are treated as two different coins in part (a) whereas they are effectively treated as the same coin in part (b). The reason they are treated as one coin is because they both contribute the same amount to the sum of money.

Read more about Counting Problems at; brainly.com/question/13875198

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3 0
2 years ago
Ask Your Teacher Consider a binomial random variable with n = 5 and p = 0.8. Let x be the number of successes in the sample. Eva
Readme [11.4K]

Answer:

0.942 is the required probability.  

Step-by-step explanation:

We are given the following in the question:

x is a binomial random variable with n = 5 and p = 0.8.

Then,

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

We have to evaluate:

P(x \geq 3) = P(x = 3) + P(x = 4) +P(x=5)\\\\= \binom{5}{3}(0.8)^3(1-0.8)^2 +\binom{5}{4}(0.8)^4(1-0.8)^1 +\binom{5}{5}(0.8)^5(1-0.8)^0\\\\= 0.2048 +0.4096+0.3276=0.942

0.942 is the required probability.

4 0
3 years ago
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