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3 years ago

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On Mother’s Day,Mr.Davis took Mrs.Davis and their daughter,Sue,to a restaurant for dinner.All three ordered the Mother’s Day Spe

cial,regularly priced at $24.80.
Mrs. Davis received a 40% discount for her dinner.
Sue’s dinner was half price for a child portion.

What was the total amount of the bill for the three meals after the discounts

1 answer:

meriva3 years ago

4 0

Answer:

$52.08

Step-by-step explanation:

Mr. Davis got the Mother's Day Special regularly priced, which is $24.80. For Mrs. Davis, 40% off of $24.80 is $14.88. For Sue, half of $24.80 is $12.40. Once you add them add up, you get $52.08.

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This week, we are covering relationships that can be approximated by linear equations. For instance, y = 453x + 3768 represents

lana [24]

Answer:

See explanation below.

Step-by-step explanation:

We assume that the data is given by :

x: 30, 30, 30, 50, 50, 50, 70,70, 70,90,90,90

y: 38, 43, 29, 32, 26, 33, 19, 27, 23, 14, 19, 21.

Where X represent the cost for scholarships in thousands of dollars and y represent the cost of life for an academic semester (The data comes from the web)

We can find the least-squares line appropriate for this data.

For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

So we can find the sums like this:

$\sum_{i=1}^n x_i = 30+30+30+50+50+50+70+70+70+90+90+90=720$

$\sum_{i=1}^n y_i =38+43+29+32+26+33+19+27+23+14+19+21=324$

$\sum_{i=1}^n x^2_i =30^2+30^2+30^2+50^2+50^2+50^2+70^2+70^2+70^2+90^2+90^2+90^2=49200$

$\sum_{i=1}^n y^2_i =38^2+43^2+29^2+32^2+26^2+33^2+19^2+27^2+23^2+14^2+19^2+21^2=9540$

$\sum_{i=1}^n x_i y_i =30*38+30*43+30*29+50*32+50*26+50*33+70*19+70*27+70*23+90*14+90*19+90*21=17540$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=49200-\frac{720^2}{12}=6000$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=17540-\frac{720*324}{12}{12}=-1900$

And the slope would be:

$m=-\frac{1900}{6000}=-0.317$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{720}{12}=60$

$\bar y= \frac{\sum y_i}{n}=\frac{324}{12}=27$

And we can find the intercept using this:

$b=\bar y -m \bar x=27-(-0.317*60)=46.02$

So the line would be given by:

$y=-0.317 x +46.02$

We have an inverse linear relationship since the slope is negative between the variables of interest.

8 0

3 years ago

Manuel borrowed $1,500 from his bank and signed a

Mice21 [21]

Answer:

The total interest to pay back on the loan is $122.40

Step-by-step explanation:

Here. we want to get the amount of interest on the loan

We are going to use the compound interest formula here

A = P(1 + r/n)^nt

where A is the amount to be paid back

P is the amount borrowed called the principal which is $1,500

r is the interest rate which is 4% annually

This is same as 4/100 = 0.04

n is the number of times the interest will be compounded per year ; annually means yearly and so for this case, n = 1

t is the number of years = 2

So we have;

A = 1500(1 + 0.04/1)^(2)(1)

A = 1500(1.04)^(2)

A =$ 1,622.40

So the amount of interest is the difference between the amount borrowed and the value to pay back

That will be;

$1,622.40 - $1,500

= $122.40

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2 years ago

Factor completely x^2-12x+36

Elenna [48]

Answer:

${(x - 6)}^{2}$

Step-by-step explanation:

$= {x}^{2} - 12x + 36 \\ = {x}^{2} - (6 + 6)x + 36 \\ = {x }^{2} - 6x - 6x + 36 \\ = x(x - 6) - 6(x - 6) \\ =(x - 6)(x - 6) \\ = {(x - 6)}^{2}$

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2 years ago

Solve each inequality, and then drag the correct solution graph to the inequality.

Nesterboy [21]

The correct solution graph to the inequalities are

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$1.6(x+8)\geq 38.4$ → B

(NOTE: The graphs are labelled A, B and C from left to right)

For the first inequality,

$4(9x-18)>3(8x+12)$

First, clear the brackets,

$36x-72>24x+36$

Then, collect like terms

$36x-24x>36+72\\12x >108$

Now divide both sides by 12

$\frac{12x}{12} > \frac{108}{12}$

∴ $x > 9$

For the second inequality

$-\frac{1}{3}(12x+6) \geq -2x +14$

First, clear the fraction by multiplying both sides by 3

$3 \times[-\frac{1}{3}(12x+6)] \geq3 \times( -2x +14)$

$-1(12x+6) \geq -6x +42$

Now, open the bracket

$-12x-6 \geq -6x +42$

Collect like terms

$-6 -42\geq -6x +12x$

$-48\geq 6x$

Divide both sides by 6

$\frac{-48}{6} \geq \frac{6x}{6}$

$-8\geq x$

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For the third inequality,

$1.6(x+8)\geq 38.4$

First, clear the brackets

$1.6x + 12.8\geq 38.4$

Collect likes terms

$1.6x \geq 38.4-12.8$

$1.6x \geq 25.6$

Divide both sides by 1.6

$\frac{1.6x}{1.6}\geq \frac{25.6}{1.6}$

∴ $x \geq 16$

Let the graphs be A, B and C from left to right

The first graph (A) shows $x\leq -8$ and this matches the 2nd inequality

The second graph (B) shows $x \geq 16$ and this matches the 3rd inequality

The third graph (C) shows $x > 9$ and this matches the 1st inequality

Hence, the correct solution graph to the inequalities are

$4(9x-18)>3(8x+12)$ → C

$-\frac{1}{3}(12x+6) \geq -2x +14$ → A

$1.6(x+8)\geq 38.4$ → B

Learn more here: brainly.com/question/17448505

8 0

2 years ago

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Scorpion4ik [409]

Answer:

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Step-by-step explanation:

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3 years ago