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3 years ago

Find the interest earned: $26,000 for 36 months at 7.5% annual compound interest

Mathematics

1 answer:

stich3 [128]3 years ago

7 0

Answer:

Interest = $6,300 (Approx.)

Step-by-step explanation:

Given:

Amount deposit (P) = $26,000

Number of month = 36 month

Number of year (n) = 36 / 12 = 3 year

Rate of interest (r) = 7.5% = 0.075

Find:

Interest earned

Computation:

Interest = p[(1+r)ⁿ-1]

Interest = 26,000[(1+0.075)³-1]

Interest = 26,000[(1.075)³-1]

Interest = 6,299.71

Interest = $6,300 (Approx.)

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Market researchers wanted to know whether the placement of a new product on a supermarket shelf significantly increases the perc

nignag [31]

Answer:

a. A two-sample z-test for a difference in sample proportions

Step-by-step explanation:

We are told that there are two supermarkets in the research named X and Y.

That a random number of shoppers were selected in each supermarket.

Thus means we have two different sample proportion.. Since a different number of people were randomly selected from each supermarket and a different number of people purchased the product.

Thus, we will have 2 sample proportions one for supermarket X and the other for supermarket Y.

In addition, the position of the product was different in both supermarkets.

Thus, we can say that this is a 2 sample Z-test for difference in their sample proportion.

4 0

3 years ago

Plot the zeros of the polynomial y = x4 + 3x3 − 27x2 + 13x + 42

a_sh-v [17]

Answer:

See explanation

Step-by-step explanation:

Consider the polynomial $y=x^4+3x^3-27x^2+13x+42$ 4th power polynomial function has at most 4 zeros.

Integer zeros can be only among the divisors of 42:

$\pm1, \ \pm 2,\ \pm 3,\ \pm 6,\ \pm 7,\ \pm 14,\ \pm 21,\ \pm 42$

Check them:

$y(1)=1^4+3\cdot 1^3-27\cdot 1^2+13\cdot 1+42=1+3-27+13+42=32\neq 0\\ \\y(-1)=(-1)^4+3\cdot (-1)^3-27\cdot (-1)^2+13\cdot (-1)+42=1-3-27-13+42=0\\ \\y(2)=2^4+3\cdot 2^3-27\cdot 2^2+13\cdot 2+42=16+24-108+26+42=0\\ \\y(3)=3^4+3\cdot 3^3-27\cdot 3^2+13\cdot 3+42=81+81-243+39+42= 0\\ \\y(-7)=(-7)^4+3\cdot (-7)^3-27\cdot (-7)^2+13\cdot (-7)+42=2,401-1,029-1,323-91+42= 0$

Thus,

$y=(x+7)(x+1)(x-2)(x-3)$

Zeros are plotted in attached diagram.

4 0

4 years ago

Read 2 more answers

Solve 7|3f + 4| = 91 for f. A. –29 or 312⁄3 B. 3 or –17⁄3 C. –3 or 17⁄3 D. –312⁄3 or 29

ryzh [129]

Answer:

Step-by-step explanation:

Given

7 | 3f + 4 | = 91 ← divide both sides by 7

| 3f + 4 | = 13

The absolute value function always returns a positive value, but the expression inside can be positive or negative. Thus there are 2 possible solutions.

Solve 3f + 4 = 13 → 3f = 13 - 4 = 9 ⇒ f = 3

Solve 3f + 4 = - 13 ⇒ 3f = - 13 - 4 = - 17 ⇒ f = - $\frac{17}{3}$

As a check

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

f = 3 : 7| 9 + 4| = 7| 13| = 7 × 13 = 91 ← True

f = - $\frac{17}{3}$

7 | - 17 + 4| = 7| - 13| = 7 × 13 = 91 ← True

Hence the correct solutions are B

4 0

4 years ago

a principal of $750 is invested in an account at 9% per year simple interest. what is the account total after 5 years

trapecia [35]

The answer is 377.5, you would times 750 by 9%. Whatever the answer to the equation is, you take the answer and times that by 5.

8 0

3 years ago

Evaluate the integral from 0 to 1/2 of xcospixdx

aleksklad [387]

$\bf \int\limits_{0}^{\frac{1}{2}}xcos(\underline{\pi x})\cdot dx\\
-----------------------------\\
u=\underline{\pi x}\implies \frac{du}{dx}=\pi \implies \frac{du}{\pi }=dx\\
-----------------------------\\
thus\implies \int\limits_{0}^{\frac{1}{2}}xcos(u)\cdot \cfrac{du}{\pi }
\\\\
\textit{wait a sec, we need the integrand in u-terms}\\
\textit{what the dickens is up with that "x"?}\qquad well\\
-----------------------------\\
u=\pi x\implies \frac{u}{\pi }=x\\
-----------------------------\\$ $\bf thus
\\\\
\int\limits_{0}^{\frac{1}{2}}\cfrac{u}{\pi }cos(u)\cdot \cfrac{du}{\pi }\implies 
\int\limits_{0}^{\frac{1}{2}}\cfrac{u}{\pi }\cdot \cfrac{cos(u)}{\pi }\cdot du\implies 
\cfrac{u}{\pi^2}\int\limits_{0}^{\frac{1}{2}}cos(u)\cdot du\\
-----------------------------\\
\textit{now, let us do the bounds}
\\\\
u\left( \frac{1}{2} \right)=\pi\left( \frac{1}{2} \right)\to \frac{\pi }{2}
\\\\
u\left( 0 \right)=\pi (0)\to 0\\
-----------------------------\\$ $\bf thus
\\\\
\cfrac{u}{\pi^2}\int\limits_{0}^{\frac{\pi }{2}}cos(u)\cdot du\implies \cfrac{u}{\pi^2}\cdot sin(u)\implies \left[ \cfrac{u\cdot sin(u)}{\pi^2} \right]_{0}^{\frac{\pi }{2}}
$

5 0

3 years ago