In a sale, there is 50% off all prices. A chair costs ?17.50 in the sale. How much was it before the sale?

1 answer:

8 0

In order to solve this problem, you double the sales price!

$17.50 x 2 = $35.00

The original price would be $35.

Hope this helps!

You might be interested in

HELP PLEASE!!!!!!!:(

NemiM [27]

Answer:

Step-by-step explanation:

$\frac{1}{2} = \frac{7}{14} \\$

$\frac{1}{7} = \frac{2}{14}$

$\frac{7}{14} + \frac{2}{14} = \frac{9}{14}$

5+2=7

$7 + \frac{9}{14} = 7 \frac{9}{14}$

3 0

2 years ago

The product of a number and four is at most -10

Ronch [10]

The product (multiply) of a number (x) and four (4) is at most (≤) -10.

4x ≤ -10

To solve this, divide both sides by 4

x ≤ -2 $\frac{1}{2}$

6 0

3 years ago

What is it called when your equation is 0x=5

laila [671]

I just write not possible

8 0

3 years ago

assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. find the

drek231 [11]

Answer:

65.3658 inches

Step-by-step explanation:

Let X be the height of a woman randomly choosen. We know tha X have a mean of 63.6 inches and a standard deviation of 2.5 inches. For an x value, the related z-score is given by z = (x-63.6)/2.5. We are looking for a value $x_{0}$ such that $P(X < x_{0}) = 0.76$ , but, $0.76 = P(X < x_{0}) = P((X-63.6)/2.5 < (x_{0}-63.6)/2.5) = P(Z < (x_{0}-63.6)/2.5)$ , i.e., $(x_{0}-63.6)/2.5$ is the 76th percentile of the standard normal distribution. So, $(x_{0}-63.6)/2.5 = 0.7063$ , $x_{0} =63.6+(2.5)(0.7063) = 65.3658$ . Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.