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Tamiku [17]
3 years ago
11

Haroldo, Xerxes, Regina, Murray, Norah, Stav, Zeke, Cam, and Georgia are invited to a dinner party. They arrive in a random orde

r and all arrive at different times. What is the probability that Xeres arrives first?
Mathematics
1 answer:
Scilla [17]3 years ago
6 0

Answer:

1/9

Step-by-step explanation:

Becuase there are 9 people invited, so they all have a 1/9 chance

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If Ryan gives 4/9 of his 36 cards to Josiah, how many cards does he give Josiah?
Tamiku [17]

Answer:

16 cards

Step-by-step explanation:

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3 years ago
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What is 36.948516 closer to 36.9 or 37
professor190 [17]
The answer will be 36.9. Hope this helps
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Q+5/9=1/6<br><br> Need help <br><br> Plz
Dennis_Churaev [7]

Answer:

q = - 7/18

Step-by-step explanation:

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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

#SPJ1

3 0
1 year ago
1. A farmer estimates her hens will produce
a_sh-v [17]

Answer:

Expected profit per dozen eggs = 4 ¢

Expected profit on the total egg production = 60,000 ¢

Step-by-step explanation:

Complete table as obtained online

Net Profit | 8 | 6 | 4 | 2 | 0 | -2

Probability |0.1|0.4|0.2|0.1|0.1|0.1

a) Expected profit per dozen eggs

Expected value is given as

E(X) = Σ xᵢpᵢ

xᵢ = each variable

pᵢ = probability of each variable

E(X) = (8×0.1) + (6×0.4) + (4×0.2) + (2×0.1) + (0×0.1) + (-2×0.1) = 4 ¢ per dozen

b) Expected profit on the total egg production

Last year, she had 12,000 dozen eggs

This year, she estimates 3,000 dozen more eggs

Estimated Total egg production this year = 12000 + 3000 = 15,000 dozen eggs

Expected profit per dozen eggs = 4 ¢

Expected profit on 15,000 dozen eggs = 15000 × 4 = 60,000 ¢

Hope this Helps!!!!

5 0
3 years ago
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