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aleksklad [387]
3 years ago
13

3 dimes, 4 pennies, 2 quarters, 1 nickel The quarters make up what part of the group of coins? write your answer as a fraction a

nd in word form.
Mathematics
1 answer:
dsp733 years ago
4 0
Add all the coins up to get a total of 10 Coins. 
2 quarters over 10 coins would be 2/10th's or 1/5ths 
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Sergio039 [100]

Answer:

x=7

Step-by-step explanation:

The two triangles are congruent so the equation is

9x-15=7x-1

9x-7x=15-1

2x=14

x=7

6 0
3 years ago
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Please help!!! This really affects my grade! I need a final answer not a guess. Thank you very much. I really appreciate it. Lin
Darina [25.2K]

Answer:

Left skewed

Step-by-step explanation:

Looking at the box and whisker plot attached showing the wait times and the number of times they occur, we can see that as we approach the left, the number of times the wait time occurs increases. The box plot also shows like as if the box was pushed to the right as the center line is closer to the right than left.

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3 years ago
One exterior angle of a regular pentagon has a measure of (2x)”. What is the value of x?
USPshnik [31]

Answer:

x = 36°

Step-by-step explanation:

The sum of the exterior angles of a pentagon is always 360°. Since the pentagon is regular, meaning that all of its sides and interior angles are congruent, we know that all of the exterior angles are also congruent. Since a pentagon has 5 sides, we know that there are 5 congruent exterior angles, therefore:

5 * 2x = 360

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4 years ago
In how many different ways can a man divide 7 different gifts among his 3 children if the eldest is to receive 3 gifts and the o
krok68 [10]

The order in which gifts are received doesn't matter - if child X gets toy 1 then toy 2, it's the same as giving child X toy 2 then toy 1 - so we are counting combinations.

The eldest child receives 3 of the 7 gifts, so they have

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possible choices of gifts.

The next child receives 2 of the remaining 4 gifts, so they have

\dbinom 42 = \dfrac{4!}{2!(4-2)!} = 6

choices.

The last child receives the remaining 2 gifts, and there is only

\dbinom22 = \dfrac{2!}{2!(2-2)!} = 1

way to select the gifts for them.

By the multiplication using, the total number of ways of distributing 7 gifts among 3 children in the prescribed way is 35 • 6 • 1 = 210.

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