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3 years ago

13

a collection of coins consist of nickels, dimes, and quarters. there are 3 fewer quarters than nickels and six more dimes than q

uarters. how many of each kind of coin is in the collection if the total value of the collection is $4.35?

1 answer:

ZanzabumX [31]3 years ago

7 0

There are 12 Nickels, 9 Quarters, and 15 Dimes.

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70

Step-by-step explanation:

26 times 2 the plus 18 and you get 70 hope this helps

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5/3

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3 years ago

What’s the answer but you also have to simplify

Jlenok [28]

Answer:

<em><u>The</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>9x</u></em><em><u>²</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>3x</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>2</u></em><em><u>.</u></em>

Step-by-step explanation:

1) Collect like terms.

$(7x^{2} + {2x}^{2} ) + ( - 2x + 5x) + (8 - 10)$

2) Simplify.

${9x}^{2} + 3x - 2$

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6 0

3 years ago

Twelve percent of U.S. residents are in their forties. Consider a group of nine U.S. residents selected at random. Find the prob

miskamm [114]

Answer:

The probability is 28 % ( approx )

Step-by-step explanation:

Since, there are only two possible outcomes in every trails ( residents are in their forties or not ),

So, this is a binomial distribution,

Binomial distribution formula,

Probability of success in x trials,

$P(x)=^nC_x p^x q^{n-x}$

Where, $^nC_x=\frac{n!}{x!(n-x)!}$

p and q are probability of success and failure respectively and n is the total number of trials.

Let X be the event of resident who are in his or her forties.

Here, p = 12% = 0.12,

⇒ q = 1 - p = 1 - 0.12 = 0.88,

n = 9,

Hence, the probability that two or three of the people in the group are in their forties = P(X=2) + P(X=3)

$^9C_2 (0.12)^2 (0.88)^{9-2}+^9C_3 (0.12)^3 (0.88)^{9-3}$

$=36(0.12)^2 (0.88)^7+84(0.12)^3 (0.88)^6$

$=0.279266611163$

$\approx 0.28$

$=28\%$

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3 years ago