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3 years ago

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I(5, 3), J(5, -3), and L(-4, 3) are three vertices of rectangle IJKL. What are the coordinates of the fourth vertex, K, of recta

ngle IJKL? A. (-4, -3) B. (5, 3) C. (5, -3) D. (-4, 3)

2 answers:

Ad libitum [116K]3 years ago

6 0

Answer:

A. $K(x,y) = (-4,-3)$

Step-by-step explanation:

On a cartesian plane, rectangle whose each pair of parallel side are parallel to respective axis is a quadrilateral must observe the following conditions:

$I(x,y) = (a, b)$ , $J(x,y) = (a, c)$ , $L(x,y) = (d, b)$ , $K(x,y) = (d, c)$

If we know that $a = 5$ , $b = 3$ , $c = -3$ , $d = -4$ , then the coordinates of the vertex K of the rectangle IJKL is:

$K(x,y) = (-4,-3)$

Hence, the correct answer is A.

Mice21 [21]3 years ago

6 0

Answer:A. -4 -3

Step-by-step explanation:

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A population of 805 cheetahs decreases by 10% per year. How many cheetahs will there be in the population after 6 years?

Phantasy [73]

Answer:

Step-by-step explanation:

y = a(b)^x

a would represent the initial value.

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x represents the number of years

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y = 427.8

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3 years ago

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nika2105 [10]

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3 years ago

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The Schuller family has five members. Dad is 6ft 2in tall. Mom is 3 inches shorter than Dad, but 2 inches taller than Ivan. Marc

ohaa [14]

Answer:

62 inches

Step-by-step explanation:

Let x be the height ( in inches ) of Ivan,

∵ Marcia is 5 inches shorter than Ivan,

height of Marcia = x - 5,

Marcia is twice as tall as Sally-Jo,

height of sally-jo = $\frac{x-5}{2}$

Mom is 2 inches taller than Ivan.

⇒ height of mom = x + 2,

Mom is 3 inches shorter than Dad,

height of dad = x + 2 + 3 = x + 5,

So, mean height of family is,

$\frac{x+x-5+\frac{x-5}{2}+x+2+x+5}{5}$

$=\frac{2x+2x-10+x-5+2x+4+2x+10}{10}$

$=\frac{9x-1}{10}$

According to the question,

x + 5 = 74 ( 1 ft = 12 in )

x = 69

Hence, mean height of the family = $\frac{9\times 69-1}{10}$

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6 0

3 years ago

"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen

Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± $Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$ ]

Where

p' is the sample proportion/point estimator of the population proportion

$Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$ is the margin of error (d) of the confidence interval.

$Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648$

So

$d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$

$d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }$

$(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}$

$n*(d*Z_{1-\alpha /2})^2= p'(1-p')$

$n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}$

$n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}$

n= 147.28 ≅ 148 boxes.

I hope it helps!

3 0

4 years ago

Multiply to -260 add to 7

djyliett [7]

What can you be more specific like a put it in a sentence. For example x*-260+7=?

6 0

4 years ago