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2 years ago

What is the selling price of an item if the original cost is $784.50 and the markup on the item is 6.5 percent?

Mathematics

1 answer:

Airida [17]2 years ago

3 0

784.5+(784.5×0.065)
=784.5+50.9925==835.4925 round the answer=835.5

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Please solve and explain the problem below

Digiron [165]

Answer:

The car would have to sell for $20,000

Step-by-step explanation:

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3 years ago

0.4(2x-9)-0.6(4x+5)=

masha68 [24]

Answer:

From what I can comprehend I believe the answer is 3.2x + -4

Step-by-step explanation:

0.4(2x-9) = 0.8x - 9

0.8x - 9 - 0.6(4x+5)=

0.8x - 9 + 2.4x+5

0.8+2.4 = 3.2x

-9 + 5 = -4

3.2x + -4

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2 years ago

Find the greatest number of six digit which is a perfect square.

Vesna [10]

A "perfect square" is an integer that is the square of an integer.

The largest 6-digit integer is 999,999.

The square root of it is 999.9995

So the greatest square of an integer is the square of 999 = 998,001 .

7 0

3 years ago

Read 2 more answers

A random variable X follows the uniform distribution with a lower limit of 670 and an upper limit of 750.a. Calculate the mean a

DENIUS [597]

You can compute both the mean and second moment directly using the density function; in this case, it's

$f_X(x)=\begin{cases}\frac1{750-670}=\frac1{80}&\text{for }670\le x\le750\\0&\text{otherwise}\end{cases}$

Then the mean (first moment) is

$E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x\,\mathrm dx=710$

and the second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x^2\,\mathrm dx=\frac{1,513,900}3$

The second moment is useful in finding the variance, which is given by

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2=\dfrac{1,513,900}3-710^2=\dfrac{1600}3$

You get the standard deviation by taking the square root of the variance, and so

$\sqrt{V[X]}=\sqrt{\dfrac{1600}3}\approx23.09$

8 0

2 years ago

(3x)(3x^)(-3x?) Simplify

Cloud [144]

Answer:

-27x³

Step-by-step explanation:

3 0

2 years ago