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4 years ago

9

Two sporting goods stores are having discount sales on basketballs at one store a basketball is on sale for 25% off the regular

price of 24.95 at the other store the same kind of basketball is on sale for 30% off the regular price of 25.80 what is the difference between the sale prices of the two stores

2 answers:

JulsSmile [24]4 years ago

6 0

Answer:

0.66

Step-by-step explanation:

24.95 x 25%=6.23

24.94 - 6.23 = 18.72

25.80 x 30% = 7.74

25.80 - 7.74 = 18.06

18.72 -18 .06 = 0.66

Tresset [83]4 years ago

3 0

For this case we have:

Store 1:

We propose the following rule of three:

24.95 -----------> 100%

x -------------------> 25%

Where the variable x represents the discount amount:

$x = \frac {25 * 24.95} {100}\\x = 6.2375$

Thus, the price of the ball is:

$24.95-6.2375 = 18.7125$

Store 2:

We propose the following rule of three:

25.80 -----------> 100%

x -------------------> 30%

Where the variable x represents the discount amount:

$x = \frac {30 * 25.80} {100}\\x = 7.74$

Thus, the price of the ball is:

25.80-7.74 = 18.06

Thus, the price difference is:

$18.7125-18.06 = 0.6525$

ANswer:

0.6525

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3 years ago

I need help finding the whole question, i'm not sure if what i've done already if correct, thanks

artcher [175]

Median of 15

(14 + b) / 2 = 15
14 + b = 15 * 2
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b = 30 - 14
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range of 30
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mean of 18
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check..
7,8,14,16,26,37
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3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago