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3 years ago

Which expression is not equivalent to 8 · x · 2 · y? 16yx 8x + 2y 16xy

Mathematics

2 answers:

fenix001 [56]3 years ago

8 0

$8x+2y$ is not equivalent to the given expression.

IceJOKER [234]3 years ago

4 0

Answer:

8x + 2y.

Step-by-step explanation:

8 . x . 2 . y = 16xy = 16yx

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PLEASE SOLVE THIS :(

Lesechka [4]

Answer:

10.5 %

Skills needed: Financial Math Essentials

Step-by-step explanation:

1) First, before getting started, let's assume the price of the product is $x$ . This variable will be used a lot throughout the problem ( $x$ ).

2) Marking a price above means increasing the price in order to make money off of the purchased product. When raising something by $y$ percent, the new price would be $x + \frac{y}{100}*x$ .

---> In this case, the price increased by $\textbf{30}$ percent.

This means that it would be: $x+\frac{30}{100}x = x+0.3x=1.3x$

New price is: $1.3x$

3) The shopkeeper is then offering a $\textbf{15}$ percent discount off of this marked price. When offering a $b$ percent discount price, the new price (with discount), expressed algebraically is: $a-\frac{b}{100}a$

---> the expression above simplifies to $(1-\frac{b}{100})a$

In this case, $a=1.3x$ , $b=15$

---> $(1-\frac{15}{100})*1.3x = (1-0.15)*1.3x=0.85*1.3x=1.105x$

This means that $x$ , with discount, has been raised $\textbf{10.5 percent}$ .

10.5 % is the profit percent

(The profit percent being the final marked up price - purchased price)

5 0

3 years ago

3 x 7/12 do not simplify pls

Vesna [10]

Hi I am here to help :). I think the answer would be 7/4. Good luck!

3 0

2 years ago

Read 2 more answers

An education researcher claims that 58% of college students work year-round. In a random sample of 400 college students, 232

Hatshy [7]

Answer:

The proportion of college students who work year-round is 58%.

Step-by-step explanation:

The claim made by the education researcher is that 58% of college students work year-round.

A random sample of 400 college students, 232 say they work year-round.

To test the researcher's claim use a one-proportion z-test.

The hypothesis can be defined as follows:

H₀: The proportion of college students who work year-round is 58%, i.e. p = 0.58.

Hₐ: The proportion of college students who work year-round is 58%, i.e. p ≠ 0.58. C

Compute the sample proportion as follows:

$\hat p=\frac{232}{400}=0.58$

Compute the test statistic value as follows:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.58-0.58}{\sqrt{\frac{0.58(1-0.58)}{400}}}=0$

The test statistic value is 0.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value for the two-tailed test as follows:

$p-value=2\times P(z$

*Use a z-table for the probability.

The p-value of the test is 1.

The p-value of the test is very large when compared to the significance level.

The null hypothesis will not be rejected.

Thus, it can be concluded that the proportion of college students who work year-round is 58%.

6 0

4 years ago

What are the x and y intercepts of 2x-3y=6?

Annette [7]

X=67/2
y=-67/3
x and y intercepts

8 0

4 years ago

Read 2 more answers

Find the future value of an annuity due of $7,000 paid at the beginning of each 6-month period for 8 years if the interest rate

Oliga [24]

Answer: Im not sure but i think its 120,960

Step-by-step explanation: i know that there is 2 of the 6-month periods in a year and multiplied that by 8 and got 16. Then i did 7000x.08=560 and then i added that to 7000 and got 7560. FInally i multiplied that by 16 and got 120,960.

4 0

3 years ago