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ale4655 [162]
3 years ago
8

Segment DE has endpoints at D(2, - 8) and E(-4, 4) . If it is dilated about the origin by a factor of 4, which of the following

would be the length of its image, D’E’
1.) 6sqrt(5);

2.) 24sqrt(5);

3.) 144sqrt(5);

4.) 10sqrt(5);

Mathematics
1 answer:
VLD [36.1K]3 years ago
3 0

Answer:

answer 2)  24\sqrt{5}

Step-by-step explanation:

after the dilation pt D is (8,-32)

after the dilation pt E is (-16,16)

if you use the distance formula for these two points you get \sqrt{2880} which simplifies to 24\sqrt{5}

You might be interested in
. upper left chamber is enlarged, the risk of heart problems is increased. The paper "Left Atrial Size Increases with Body Mass
Sonbull [250]

Answer:

Part 1

(a) 0.28434

(b) 0.43441

(c) 29.9 mm

Part 2

(a) 0.97722

Step-by-step explanation:

There are two questions here. We'll break them into two.

Part 1.

This is a normal distribution problem healthy children having the size of their left atrial diameters normally distributed with

Mean = μ = 26.4 mm

Standard deviation = σ = 4.2 mm

a) proportion of healthy children have left atrial diameters less than 24 mm

P(x < 24)

We first normalize/standardize 24 mm

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (24 - 26.4)/4.2 = -0.57

The required probability

P(x < 24) = P(z < -0.57)

We'll use data from the normal probability table for these probabilities

P(x < 24) = P(z < -0.57) = 0.28434

b) proportion of healthy children have left atrial diameters between 25 and 30 mm

P(25 < x < 30)

We first normalize/standardize 25 mm and 30 mm

For 25 mm

z = (x - μ)/σ = (25 - 26.4)/4.2 = -0.33

For 30 mm

z = (x - μ)/σ = (30 - 26.4)/4.2 = 0.86

The required probability

P(25 < x < 30) = P(-0.33 < z < 0.86)

We'll use data from the normal probability table for these probabilities

P(25 < x < 30) = P(-0.33 < z < 0.86)

= P(z < 0.86) - P(z < -0.33)

= 0.80511 - 0.37070 = 0.43441

c) For healthy children, what is the value for which only about 20% have a larger left atrial diameter.

Let the value be x' and its z-score be z'

P(x > x') = P(z > z') = 20% = 0.20

P(z > z') = 1 - P(z ≤ z') = 0.20

P(z ≤ z') = 0.80

Using normal distribution tables

z' = 0.842

z' = (x' - μ)/σ

0.842 = (x' - 26.4)/4.2

x' = 29.9364 = 29.9 mm

Part 2

Population mean = μ = 65 mm

Population Standard deviation = σ = 5 mm

The central limit theory explains that the sampling distribution extracted from this distribution will approximate a normal distribution with

Sample mean = Population mean

¯x = μₓ = μ = 65 mm

Standard deviation of the distribution of sample means = σₓ = (σ/√n)

where n = Sample size = 100

σₓ = (5/√100) = 0.5 mm

So, probability that the sample mean distance ¯x for these 100 will be between 64 and 67 mm = P(64 < x < 67)

We first normalize/standardize 64 mm and 67 mm

For 64 mm

z = (x - μ)/σ = (64 - 65)/0.5 = -2.00

For 67 mm

z = (x - μ)/σ = (67 - 65)/0.5 = 4.00

The required probability

P(64 < x < 67) = P(-2.00 < z < 4.00)

We'll use data from the normal probability table for these probabilities

P(64 < x < 67) = P(-2.00 < z < 4.00)

= P(z < 4.00) - P(z < -2.00)

= 0.99997 - 0.02275 = 0.97722

Hope this Helps!!!

7 0
4 years ago
3x+2=11 plz hep im slow in math
stealth61 [152]

Answer:

3

Step-by-step explanation:

3×3=9+2=11. so 3 is the answer.

3 0
3 years ago
On Martin's first stroke, his golf ball traveled 4/5 of the distance to the hole. On his second stroke, the ball traveled 79 met
Sav [38]

Answer:

0.395 kilometre

Step-by-step explanation:

Given:

On Martin's first stroke, his golf ball traveled 4/5 of the distance to the hole.

On his second stroke, the ball traveled 79 meters and went into the hole.

<u>Question asked:</u>

How many kilometres from the hole was Martin when he started?

<u>Solution:</u>

Let distance from Martin starting point to the hole in meters = x

On Martin's first stroke, ball traveled = \frac{4}{5} \ of \ total \ distance\ to\ the\ hole

                                                             =\frac{4}{5} \times x=\frac{4x}{5}

On his second stroke, the ball traveled and went to the hole = 79 meters

Total distance from starting point to the hole = Ball traveled from first stroke + Ball traveled from second stroke

x=\frac{4x}{5} +79\\ \\ Subtracting\ both\ sides\ by \ \frac{4x}{5}\\ \\ x- \frac{4x}{5}= \frac{4x}{5}- \frac{4x}{5}+79\\ \\ \frac{5x-4x}{5} =79\\ \\ By \ cross\ multiplication\\ \\ x=79\times5\\ \\ x=395\ meters

Now, convert it into kilometre:

1000 meter = 1 km

1 meter = \frac{1}{1000}

395 meters = \frac{1}{1000}\times395=0.395\ kilometre

Thus, there are 0.395 kilometre distance from Martin starting point to the hole.

8 0
3 years ago
Jaxon is creating a mosaic design on a rectangular
olasank [31]

Answer:

hard question

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What is 24tenths÷8
Taya2010 [7]

Answer:

.03

Step-by-step explanation:

The answer would be .03 as you are dividing .24 by 8, which would be 3 if there was no decimal point. Though, .24 is two places past the ones place so the answer would have to be two places past the ones place. If you divide it, 8 can't go into 2, so you would put a 0 above it. Though, 8 can go into 24 three times, so you would put three. Therefore, your answer would be .03 as it is two places past the ones place and if it is multiplied by 8, it would make .24, or 24 tenths.

7 0
3 years ago
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