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Alina [70]
3 years ago
7

Please help ! if i get an answer now i’ll also know how to do it in the future c:

Mathematics
1 answer:
leonid [27]3 years ago
6 0

Do they give you any are or perimeter?

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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
2 years ago
If the median of the score distribution on this pop quiz is more than 170, the teacher will drop the next pop quiz. The pop quiz
antiseptic1488 [7]

Answer:

174.50

Step-by-step explanation:

first order the numbers from least to greatest then find the middle digit. if there are two, add them together, then divide them by 2. if there's only one, that is your median.

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3 years ago
A cone has a base radius of 7 centimeters and a height of 21 centimeters.
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V = (1/3) x πr² x h = cm³
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3 years ago
The sum of two numbers is 5.036. One number has a 4 in the tenths place and a 7 in the thousandths place. The other number has a
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3.407
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These make 5.036.
5 0
4 years ago
Given that f(x)= x2+3 and g(x)= x+4/3 solve for (f(g(x)) when x =2
Savatey [412]
\bf \begin{cases}&#10;f(x)=x^2+3\\\\&#10;g(x)=x+\cfrac{4}{3}&#10;\end{cases}&#10;\\\\\\&#10;\boxed{x=2}\qquad g(2)=(2)+\cfrac{4}{3}\implies \boxed{g(2)=\cfrac{10}{3}}&#10;\\\\\\&#10;f(~~g(x)~~)=[g(x)]^2+3\implies f(~~g(2)~~)=[g(2)]^2+3&#10;\\\\\\&#10;f(~~g(2)~~)=\left[ \boxed{\frac{10}{3}} \right]^2+3\implies f(~~g(2)~~)=\cfrac{10^2}{3^2}+3&#10;\\\\\\&#10;f(~~g(2)~~)=\cfrac{100}{9}+3\implies f(~~g(2)~~)=\cfrac{100+27}{9}&#10;\\\\\\&#10;f(~~g(2)~~)=\cfrac{127}{9}
7 0
3 years ago
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