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3 years ago

11

Suppose you have 20 coins that total $4.some coins are nickles and sime are quarters .which of the following pairs of equations

would you use to find out how many of each coin you have?

1 answer:

Monica [59]3 years ago

4 0

So 1 nickle=5 cents
1 quarter=25 sents

q=number of quarters
n=number of nickles so

total is
q+n=20
5n+25q=400 (since 100 cents=1 dollar and 4 dollares=400 cents)

SO THE 2 EQUATIONS WOULD BE
n+q=20
5n+25q=400

so q+n=20
subtract q from both sides
n=20-q
subsitute 20-q for n in the second equation
5(20-q)+25q=400
distribute
100-5q+25q=400
add like terms
100+20q=400
subtract 100 from both sides
20q=300
divide both sides by 20
q=15

thhere are 15 quarters

subsitute 15 for q in first equation

15+n=20
subtract 15 from both sides
n=5

5 nickles
15 quarters

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See comment for complete question

The question is an illustration of binomial probability where

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$(a):\ P(x \ge 5)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

Using the complement rule, we have:

$P(x \ge 5) = 1 - P(x < 5)$

So, we have:

$P(x < 5) =P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)$

$P(x = 0) = ^{15}C_{0} * (75\%)^0 * (1 - 75\%)^{15 - 0} = 1 * 1 * (0.25)^{15} = 9.31 * 10^{-10}$

$P(x = 1) = ^{15}C_{1} * (75\%)^1 * (1 - 75\%)^{15 - 1} = 15* (0.75)^1 * (0.25)^{14} = 4.19 * 10^{-8}$

$P(x = 2) = ^{15}C_{2} * (75\%)^2 * (1 - 75\%)^{15 - 2} = 105* (0.75)^2 * (0.25)^{13} = 8.80 * 10^{-7}$

$P(x = 3) = ^{15}C_{3} * (75\%)^3 * (1 - 75\%)^{15 - 3} = 455* (0.75)^2 * (0.25)^{12} = 0.0000153$

$P(x = 4) = ^{15}C_{4} * (75\%)^4 * (1 - 75\%)^{15 - 4} = 1365 * (0.75)^4 * (0.25)^{11} = 0.000103$

So, we have:

$P(x < 5) = (9.31 * 10^{-10}) + (4.19 * 10^{-8}) + (8.80 * 10^{-7}) + 0.0000153 + 0.000103$

$P(x < 5) = 0.00011922283$

Recall that:

$P(x \ge 5) = 1 - P(x < 5)$

$P(x \ge 5) = 1 - 0.00011922283$

$P(x \ge 5) = 0.9998$

$P(x \ge 5) = 1.000$ --- approximated

$(b)\ P(x = 15)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x=15) = ^{15}C_{15} * 0.75^{15} * (1 - 0.75)^{15-15}$

$P(x=15) = 1 * 0.75^{15} * (0.25)^{0$

$P(x=15) = 0.013$

$(c)\ P(x = 0)$

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$P(x = 0) = ^{5}C_0 * (70\%)^0 * (1 - 70\%)^{5-0}$

$P(x = 0) = 1 * 1 * (0.30)^5$

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