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2 years ago

If a man earns $3580 per month and spends 65% of it. How much does he save every month?

2 answers:

8 0

He saves $1253 per month.

Explanation: 65% of $3580 is $2327, meaning he spends $2327 per month. He has $1253 left over because $3580-$2327=$1253.

pshichka [43]2 years ago

5 0

Answer:

$1253

Step-by-step explanation:

3580-65%=1253

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Answer:

0.0135 = 1.35% probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.

Step-by-step explanation:

For each visitor of the website, there are only two possible outcomes. Either they are looking for the website, or they are not. The probability of a customer being looking for the website is independent of other customers. 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.

5% of all visitors to the website are looking for other websites.

So 100 - 5 = 95% are looking for the website, which means that $p = 0.95$

Find the probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.

This is P(X = 2) when n = 4. So

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

$P(X = x) = C_{4,2}.(0.95)^{2}.(0.05)^{2} = 0.0135$

0.0135 = 1.35% probability that, in a random sample of 4 visitors to the website, exactly 2 actually are looking for the website.

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2 years ago

Airen’s grandparents deposited $1300 in a mutual fund earning 6% interest compounded annually. Write an equation to represent ho

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$~~~~~~ \textit{Compound Interest Earned Amount \underline{in 18 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$1300\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &18 \end{cases}$

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