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2 years ago

12

Kasey had to make a down payment of 17% of the selling price of her house. She was approved for a $255,000 mortgage. What range

of costs might she expect to pay at the closing?

1 answer:

defon2 years ago

5 0

Answer:

aq lang to Yung d na binalikan Yung iniwan na parang basura

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Find the area of this composite figure.

Margaret [11]

Answer: ≈ 196.53 m²

All work is shown in the picture.

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3 years ago

What is (5y+6)-(10y+3)

melisa1 [442]

$5y + 6 - 10y - 3 \\ - 5x + 3$

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3 years ago

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What is the radius of a circle whose equation is x2 y2 8x – 6y 21 = 0?

andrew11 [14]

we know that

the standard form of the equation of the circle is

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

$x^{2} +y^{2} +8x-6y+21=0$

<u>Convert to standard form</u>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+8x) +(y^{2}-6y)=-21$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+8x+16) +(y^{2}-6y+9)=-21+16+9$

$(x^{2}+8x+16) +(y^{2}-6y+9)=4$

Rewrite as perfect squares

$(x+4)^{2} +(y-3)^{2}=4$

$(x+4)^{2} +(y-3)^{2}=2^{2}$

the center of the circle is $(-4,3)$

the radius of the circle is $2\ units$

therefore

<u>the answer is</u>

the radius of the circle is $2\ units$

8 0

3 years ago

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Study the following data set.

Vera_Pavlovna [14]

Answer:

8

Step-by-step explanation:

The question is on interquartile range which is the median of the upper half of the data minus the median of the lower half of data

First arrange the data in an increasing order;

8,15,9,18,9,17,22,10,11,9,13

8,9,9,9,10,11,13,15,17,18,22

Find the median, which is the center value in the data set

8,9,9,9,10,11,13,15,17,18,22⇒the median is 11

Place brackets around the numbers above and below the median

(8,9,9,9,10)11 (13,15,17,18,22)

Find the median in the lower half of the data,Q1

(8,9,9,9,10) ⇒median is 9=Q1

Find the median in the upper half of the data,Q3

(13,15,17,18,22)⇒median is 17=Q3

Subtract Q1 from Q3 to get the interquartile range

Q3=17, and Q1=9

Q3-Q1=17-9=8

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3 years ago

an astronomer wishes to measure on a photograph the distance between the image of a certain star and three star other star image

Greeley [361]

The astronomer has to choose 3 star images out of 8.This is a problem of combinations. This can be expressed as 8C3 i.e. combination of 8 objects taken 3 at a time.

$8C3= \frac{8!}{3!*5!}=56$

This means, if 8 images are nearby the astronomer will have 56 choices <span>of the three stars.</span>

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3 years ago