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3 years ago

5

If 12,025 was invested at 3.25% compounded monthly 3 years, how much interest will be earned

1 answer:

wariber [46]3 years ago

6 0

Answer:

$1,229.75

Step-by-step explanation:

Lets use the compound interest formula provided to solve this:

$A=P(1+\frac{r}{n} )^{nt}$

<em>P = initial balance</em>

<em>r = interest rate (decimal)</em>

<em>n = number of times compounded annually</em>

<em>t = time</em>

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First, change 3.25% into a decimal:

3.25% -> $\frac{3.25}{100}$ -> 0.0325

Since the interest is compounded monthly, we will use 12 for n. Lets plug in the values now:

$A=12,025(1+\frac{0.0325}{12})^{12(3)}$

$A=13,254.75$

Lastly, subtract A from P to get the interest earned:

$13,254.75 - 12,025 = 1,229.75$

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3 years ago

There are 1200 students at a school.

const2013 [10]

Answer:

The number of boys that have school dinners is 381.

Step-by-step explanation:

First, we need to find how many students have school dinners:

$SD = 1200*0.62 = 744$

The percent of girls that have school dinners:

$\%SD_{g} = 100\% - 45\% = 55\%$

Now, we need to find the number of boys (b) and girls (g) at a school:

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Hence. the percent of boys that have school dinners is given by:

$SD = g*\%SD_{g} + b*\%SD_{b}$

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Finally, the number of boys that have school dinners is:

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Therefore, the number of boys that have school dinners is 381.

I hope it helps you!

4 0

3 years ago

in a given year, the rate of flu infection for the general public was 8.3%. And sample of 200 people who receive the flu vaccine

makkiz [27]

Answer:

$\fbox{\begin{minipage}{10em}Option A is correct\end{minipage}}$

Step-by-step explanation:

<em>Step 1: Define significance level</em>

In this hypothesis testing problem, significance levels α is selected: $0.05$ , the associated z-value from Laplace table:

Φ( $z$ ) = α - $0.5 = 0.05 - 0.5 = -0.45$

=> $z$ = $-1.645$

<em>Step 2: Define null hypothesis (</em> $H_{0}$ <em>) and alternative hypothesis (</em> $H_{1}$ <em>)</em>

$H_{0}$ : rate of flu infection $p$ = 8.3% or 8.3/100 = 0.083

$H_{1}$ : rate of flu infection $p$ < 8.3% or 8.3/100 = 0.083

<em>Step 3: Apply the formula to check test statistic:</em>

$K = \frac{f - p}{\sqrt{p(1 - p)} } * \sqrt{n}$

with $f$ is actual sampling percent, $p$ is rate of flu infection of $H_{0}$ , $n$ is number of samples.

The null hypothesis will be rejected if $K < z$

<em>Step 4: Calculate the value of K and compare with </em> $z$

$K = \frac{(\frac{3.5}{100}) - 0.083}{\sqrt{0.083(1 - 0.083)} } * \sqrt{200} = -2.46$

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=>This is good evidence to reject null hypothesis.

=> The actual rate is lower. (As $H_{1}$ states)

Hope this helps!

:)

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