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Andrei [34K]
3 years ago
9

Ernest and Gene have been saving coins every day for a sunny day in Arizona. Ernest has 8 more half dollars than Gene and no qua

rters. Gene has as many quarters as Ernest has half dollars, and each has 45 dimes. How many of each coin has each saved for the big day when $200 has been saved? Which of the following equations could represent the word problem if x is the number of half dollars that Gene has?a. 3x + 106 = 200
b. 125x + 916 = 20,000
c. 125x + 1,500 = 20,000
Mathematics
1 answer:
IgorC [24]3 years ago
3 0

Answer:

1. Ernest has $82.5 and Gene has $117.5

2. c) 125x + 1500 = $20000

Step-by-step explanation:

If we mark Gene's half dollars with x, then it is given that Ernest has 8 more, which is x+8.

It is also given that Ernest has no quarters, but Gene has them as many as Ernest has half dollars, which is x+8.

And it is given that they both have 45 dimes each, which is $4.5 each.

Now, let's add up all these numbers:

Ernest: (x+8)•0.5 + 0•0.25 + 4.5 = 0.5x + 8.5

Gene: x•0.5 + (x+8)•0.25 + 4.5 = 0.75x + 6.5

They collected $200 together which means:

0.5x + 8.5 + 0.75x + 6.5 = $200

1.25x + 15 = $200

If we want to avoid decimal number, we can multiply whole equation with 100:

125x + 1500 = $20000

so, the correct answer is C.

Finally, to find x:

125x = 18500

x = 148

Ernest had 0.5x + 8.5 which is $82.5

Gene had 0.75x + 6.5 which is $117.5

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Here the sum = 9000°, thus

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4. Two cards are selected from a regular deck of cards. If the first was a ten, determine the probability that the second was a
ioda

Answer:

a) P=\frac{1}{221}

b) P=\frac{4}{663}

c) P=\frac{4}{221}

Step-by-step explanation:

Given : Two cards are selected from a regular deck of cards. If the first was a ten.

To determine : The probability that the second was a (a) ten, (b) jack, (c) picture card ?

Solution :

Total number of cards = 52 in which there are 4 sets of 13 cards with two different color.

We are drawing two card one after another without replacement.

Probability of drawn first card 10 is P(T_1)=\frac{4}{52}=\frac{1}{13}

a) Second card drawn is 10,

Probability of drawn second card 10 is P(T_2)=\frac{3}{51}=\frac{1}{17}

Total probability that first card was 10 and second card is 10,

P=P(T_1)\times P(T_2)

P=\frac{1}{13}\times \frac{1}{17}

P=\frac{1}{221}

b) Second card drawn is Jack,

Probability of drawn second card Jack is P(J)=\frac{4}{51}

Total probability that first card was 10 and second card is Jack,

P=P(T_1)\times P(J)

P=\frac{1}{13}\times \frac{4}{51}

P=\frac{4}{663}

c) Second card drawn is picture card,

Number of picture card is 4j's, 4Q's, 4K's = 12

Probability of drawn second card picture card is P(p)=\frac{12}{51}=\frac{4}{17}

Total probability that first card was 10 and second card is picture card,

P=P(T_1)\times P(p)

P=\frac{1}{13}\times \frac{4}{17}

P=\frac{4}{221}

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3 years ago
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