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faust18 [17]
3 years ago
6

From 1960 to 1970, the consumer price index (CPI) increased from 29.6 to 38.8. If a dozen tangerines cost $0.31 in 1960 and the

price of tangerines increased at the same rate as the CPI from 1960 to 1970, approximately how much did a dozen tangerines cost in 1970?
Mathematics
2 answers:
trasher [3.6K]3 years ago
6 0

The percent increase in the price index from 1960 to 1970 is:

% increase in price index = (38.8 – 29.6) * 100% / 29.6

% increase in price index = 31.08%

Therefore the price of a dozen tangerines in 1970 (X) is:

31.08 = (X - $0.31) * 100 / $0.31

X = $0.41


patriot [66]3 years ago
4 0

Answer: 1.45

Step-by-step explanation:

29.6=42.898

Divide 42.898 by 29.6 and that will round up to be 1.45 which is correct on apec

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Answer:

Jamie. by 7/16

Step-by-step explanation:

Halverson completed: 3/16

Jamie completed: 10/16.

To understand who completed more, look at the denominator. The common denominator is 16. Assume a puzzle has 16 pieces and Halverson did 3 out of 16 and Jamie did 10 out of 16. This means Jamie did more.

If you want to know by how much Jamie did more, it would be the difference of both the fraction. This means that 10 done by Jamie minus 3 done by Halverson, which is 7. So 7 out of 16 is the difference.

\frac{10}{16} -\frac{3}{16}

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Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> =  Mike will show up.

Given:

P(D^{c})=0.55\\P(M^{c})=0.45

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

                                                   =1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25

Thus, the probability that neither Dave nor Mike will show up is 0.25.

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