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3 years ago

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Addison is going to invest in an account paying an interest rate of 3.1% compounded continuously. How much would Addison need to

invest, to the nearest dollar, for the value of the account to reach $910 in 18 years?

1 answer:

professor190 [17]3 years ago

7 0

Answer: $521

Step-by-step explanation:

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In a recent year, there was about 400,000,000 mobile internet users. ( look at the picture )

AVprozaik [17]

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3 : 1, or 3 to 1, thus 3 times as many.

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Keli has 26 roses and 73 carnations. She will put an assortment of 6 flowers in each vase. How many vases will she need for her

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3 years ago

Which equation have the same value of X as 3/5 ( 30x -15) = 72 Select 3 options A. 18x - 15 = 72 B. 50x -25=72 C. 18x -9= 72 D.

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18x - 9 = 72 are the equations

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3 years ago

The product of 30 degrees and 60 degrees is the same as the product of

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2 years ago

A football team has a probability of .75 of winning when playing any of the other four teams in its conference. If the games are

Alexeev081 [22]

Answer:

0.3164 = 31.64% probability the team wins all its conference games

Step-by-step explanation:

For each conference game, there are only two possible outcomes. Either the team wins it, or they lose. The probability of winning a game is independent of any other game. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

A football team has a probability of .75 of winning when playing any of the other four teams in its conference.

The probability means that $p = 0.75$ , and four games means that $n = 4$

If the games are independent, what is the probability the team wins all its conference games?

This is P(X = 4). So

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

$P(X = 4) = C_{4,4}.(0.75)^{4}.(0.25)^{0} = 0.3164$

0.3164 = 31.64% probability the team wins all its conference games

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3 years ago