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kirza4 [7]
3 years ago
15

What is the approximate area, in square units, or circle C

Mathematics
1 answer:
Anestetic [448]3 years ago
7 0
See picture hope it helps

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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
HELP MEEEEEEEEEEEEE!!!!!!!!!!!!!!!!
aliina [53]

Answer:

linear or Quadratic

i hope it works

8 0
2 years ago
Read 2 more answers
CAN SOMEONE HELP ME PLEASE??? I.DK HOW TO DO THIS :'(
antiseptic1488 [7]
B hice this Yelp you
5 0
2 years ago
What’s the correct answer
SVETLANKA909090 [29]

Answer: C. 1998, 2000, 2002

7 0
3 years ago
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Find the following limit or state that it does not exist. ModifyingBelow lim With x right arrow minus 2 StartFraction 3 (2 x min
leva [86]

Answer:

-60

Step-by-step explanation:

The objective is to state whether or not the following limit exists

                                \lim_{x \to -2}  \frac{3(2x-1)^2 - 75}{x+2}.

First, we simplify the expression in the numerator of the fraction.

3(2x-1)^2 -75 = 3(4x^2 - 4x +1) -75 = 12x^2 - 12x + 3 - 75 = 12x^2 - 12x -72

Now, we obtain

                         12(x^2-x-6) = 12(x+2)(x-3)

and the fraction is transformed into

                       \frac{3(2x-1)^2 - 75}{x+2} =  \frac{12(x+2)(x-3)}{x+2} = 12 (x-3)

Therefore, the following limit is

       \lim_{x \to -2}  \frac{3(2x-1)^2 - 75}{x+2} = \lim_{x \to -2}  12(x-3) = 12 \lim_{x \to -2} (x-3)

You can plug in -2 in the equation, hence

                        12 \lim_{x \to -2} (x-3) = 12 (-2-3) = -60

6 0
3 years ago
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