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3 years ago

You determine that the standard deviation for a sample of test scores is 0. This tells you that:

Mathematics

1 answer:

denis-greek [22]3 years ago

3 0

Answer:

Your answer is A,B,AND D

Step-by-step explanation:

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A department store buys 200 shirts at a cost of $5,400 and sells them at a selling

jekas [21]

Number of shirts a department store bought = 200

Cost price of 200 shirts = $5400

Cost price of 1 shirt :

$=\tt \frac{5400}{200}$

$\color{plum} =\tt \$27$

Thus, the cost price of one shirt = $27

Price at which the department store sold each shirt = $30

Which means :

▪︎Cost price of shirt = $27

▪︎Selling price of shrit = $30

Since selling price is greater than cost price, there was a markup in the price of the shirts.

We know that :

$\color{hotpink}\tt \: markup \: percentage = \color{plum} \frac{selling \: price - cost \: price}{cost \: price } \times 100$

Markup percentage in the price of the shirts :

$= \tt \frac{30 - 27}{27} \times 100$

$= \tt \frac{3}{27} \times 100$

$= \tt \frac{3 \times 100}{27}$

$= \tt \frac{300}{27}$

$\color{plum} = \tt11.1 \%$

Thus, the percentage of markup = $\tt 11.1 \%$

Therefore, the percentage of markup in the shirt's price = $\color{hotpink} \tt11.1 \%$

3 0

3 years ago

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The height of an object varies directly with the length of its shadow, A person who is 6 feet tall casts a 15-foot shadow, how l

11Alexandr11 [23.1K]

The multiplier to get from 6 * ... to 15 is 15/6.
If 6 * 15/6 = 15, then 20 * 15/6 = 50, answer B.

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3 years ago

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

IRINA_888 [86]

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

$A=1300\left(1+\frac{0.06}{1}\right)^{1\cdot 18}\implies A=1300(1.06)^{18}\implies A\approx 3710.64 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y = 1300(1.06)^x~\hfill$