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3 years ago

14

A local neighborhood store reduced its price for 32" tv sets 10% from the original price of $325.

1 answer:

LUCKY_DIMON [66]3 years ago

5 0

Answer:

$32.50

Step-by-step explanation:

100 - 10 = 90%

sale price = 90%

325/100 X 90 = 292.50

325 - 292.50 = 32.50

discount = $32.50

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Find two numbers, if their sum is −11 and their difference is 41

snow_tiger [21]

The larger one is (-11 + 41)/2 = 15.

The smaller one is (-11 -41)/2 = -26.

_____

For two numbers a and b with sum s and difference d, you can write the equations

$a+b=s\\a-b=d$

Then adding the equations and dividing that sum by 2, you get

$(a+b)+(a-b)=(s)+(d)\\2a=s+d\\a=\dfrac{s+d}{2}$

You can subtract the second eqution from the first and get a similar result for the smaller number

$(a+b)-(a-b)=(s)-(d)\\2b=s-d\\b=\dfrac{s-d}{2}$

These are the formulas we used above.

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4 years ago

What's the square root of 100?

Assoli18 [71]

10 hope it helps you out

3 0

3 years ago

Read 2 more answers

Pre-Calculus - Systems of Equations with 3 Variables please show work/steps

Tatiana [17]

Answer:

x = 10 , y = -7 , z = 1

Step-by-step explanation:

Solve the following system:

{x - 3 z = 7 | (equation 1)

2 x + y - 2 z = 11 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Swap equation 1 with equation 2:

{2 x + y - 2 z = 11 | (equation 1)

x + 0 y - 3 z = 7 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y/2 - 2 z = 3/2 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Multiply equation 2 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Add 1/2 × (equation 1) to equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - (3 y)/2 + 8 z = 37/2 | (equation 3)

Multiply equation 3 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - 3 y + 16 z = 37 | (equation 3)

Swap equation 2 with equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x - y - 4 z = 3 | (equation 3)

Subtract 1/3 × (equation 2) from equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y - (28 z)/3 = (-28)/3 | (equation 3)

Multiply equation 3 by -3/28:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 16 × (equation 3) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y+0 z = 21 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 2 by -3:

{2 x + y - 2 z = 11 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{2 x + 0 y - 2 z = 18 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Add 2 × (equation 3) to equation 1:

{2 x+0 y+0 z = 20 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by 2:

{x+0 y+0 z = 10 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer: {x = 10 , y = -7 , z = 1

3 0

3 years ago

Find 3 ordered pairs which are Solutions to the equation 2x - 3y = 12

KonstantinChe [14]

2x-3(0)=12
2x=12
x=6
(6,0)

2(0)-3y=12
-3y=12
y=-4
(0,-4)

2x-3(2)=12
2x-6=12
2x=18
x=9
(9,2)

ANSWERS: (6,0) (0,-4) (9,2)

7 0

4 years ago

Noise levels at 3 airports were measured in decibels yielding the following data:

Shalnov [3]

Answer:

a) The 90% confidence interval for the mean noise level at such locations

(112.46 , 163.54)

b) The critical value that should be used in constructing the confidence interval

Z₀.₁₀ = 1.645

Step-by-step explanation:

<u><em>Step(i)</em></u> :-

Noise levels at 3 airports were measured in decibels yielding the following data

108 146 160

Mean of given data

$x^{-} = \frac{108+146+160}{3} = 138$

x : 108 146 160

x-x⁻ : -30 8 22

(x-x⁻ )² : 900 64 484

Variance σ ² = ∑(x-x⁻ )²/ n-1

= $\frac{900+64+484}{3-1}= 724$

Standard deviation

σ = √724 = 26.90

<u><em>Step(ii):</em></u>-

The 90% confidence interval for the mean noise level at such locations

$(x^{-} - Z_{0.90} \frac{S.D}{\sqrt{n} } , (x^{-} + Z_{0.90} \frac{S.D}{\sqrt{n} } )$

The critical value that should be used in constructing the confidence interval

Z₀.₁₀ = 1.645

$(138 - 1.645 \frac{26.90}{\sqrt{3} } , (138 + 1.645 \frac{26.90}{\sqrt{3} } )$

( 138 - 25.54 , 138 + 25.54 )

(112.46 , 163.54)

4 0

3 years ago