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11 months ago

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the life expectancy of a circulating coin is 30 years. the life expectancy of a circulating dollar bill is only 1/20 as long. fi

nd the life expectancy of circulating paper money. Life expectancy= how many years ( type a whole number or a mixed number simplify your answer)

1 answer:

Flauer [41]11 months ago

4 0

Given

the life expectancy of a circulating coin is 30 years

the life expectancy of a circulating dollar bill is only 1/20 as long.

Procedure

c=life expectancy of a circulating coin

p= life expectancy of circulating paper money

$p=\frac{1}{20}\cdot c$ $\begin{gathered} p=\frac{1}{20}\cdot30 \\ p=\frac{30}{20} \\ p=1.5 \end{gathered}$

The answer is: life expectancy of a dollar bill = 1.5 years

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Match the types of well formed formulas with the expressions. Negation Conjunction Disjunction Conditional Biconditional A. T &l

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<h3>Answers:</h3>

A. T <-> U is a <u>biconditional</u>
B. (A & B) v (C & D) is a <u>disjunction</u>
C. R -> ~S is a <u>conditional</u>
D. P & Q is a <u>conjunction</u>
E. ~(R v P) is a <u>negation</u>

========================================

Explanations:

A biconditional is anything in the form A <-> B. This is a compact way of saying (A -> B) & (B -> A). We replace A and B with logical statements.
Disjunctions are of the basic form A v B. The "v" basically means "or".
Any conditional is of the form "if... then...". For example, "if it rains, then it gets wet outside" is a conditional. In terms of logic symbols, we write A -> B to mean "if A, then B".
Conjunctions are whenever we combine two logical statements with an "and" or an ampersand symbol. The basic form is A & B
Negations are the complete opposite of the original. If the original is P, then the negation is ~P, which is read as "not P".

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3 years ago

There are fifteen teams in a high school baseball league. How many different orders of finish are possible for the first four p

ankoles [38]

There are 32,760 different orders for the first four positions.

This is a permutation where repetition is not allowed; the formula for that is:

$_nP_r=\frac{n!}{(n-r)!}$

For our problem, we have:
$_{15}P_4=\frac{15!}{(15-4)!}=\frac{15!}{11!}=32760$

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3 years ago

If 2+2 is 4 then who ended up going to Adrian's kickback?

Naily [24]

Answer:

Uneducated people.

Step-by-step explanation:

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2 years ago

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A farmer packed 3 pints of strawberries every 4 minutes. In the afternoon she packed 2 pints of strawberries every 3 minutes. Wh

MakcuM [25]

Answer:

5 (strawberries / hours)

Step-by-step explanation:

calculation fro morning

strawberries / minutes x minutes / hours = strawberries / hours

so after adding the value in above equation

3/4* 60/1 = 45 strawberries / hours

calculation in the afternoon

strawberries / minutes x minutes / hours = strawberries / hours

2/3 x 60/1 = 40 strawberries / hours

so now by calculating difference between morning and afternoon packing rates, you can easily calculate

45-40 = 5 (strawberries / hours)

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2 years ago

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

So

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago