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Lapatulllka [165]
2 years ago
15

Which sentence describes why polygon MNOP is congruent to polygon JKLP?

Mathematics
2 answers:
alukav5142 [94]2 years ago
5 0

Answer:

B

Step-by-step explanation:

bogdanovich [222]2 years ago
3 0

Answer:

Option B.

Polygon JKLP maps to polygon MNOP through a rotation.

Step-by-step explanation:

The picture of the question in the attached figure

we know that

If two figures are congruent, then its corresponding sides and its corresponding angles are congruent

In this problem

JK≅MN

KL≅NO

LP≅OP

JP≅MP

and

∠OPM≅∠LPJ

∠NOP≅∠KLP

∠ONM≅∠LKJ

∠NMP≅∠KJP

so

Polygon JKLP maps to polygon MNOP through a rotation clockwise around point P an angle equal to ∠JPM

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mamaluj [8]
So the new total is:
9 * (44*.8) since the 20% discount means you only pay 80% of the original $44 per item
4 0
2 years ago
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The combined average weight of an okapi and a llama is 450450450 kilograms. The average weight of 333 llamas is 190190190 kilogr
Kisachek [45]
I'm going to assume that you meant 450kg for the combined weight, 190kg more and 3 Llamas. I'm pretty sure Llamas and Okapis don't weigh 450450450kg (that's 993,073,252 pounds). :)

x= Okapi weight
y= Llama weight

EQUATIONS:
There are 2 equations to be written:

1) 450kg is equal to the weight of one Okapi and one Llama

450kg= x + y

2) The weight of 3 llamas is equal to the weight of one Okapi plus 190kg.

3y=190kg + x


STEP 1:
Solve for one variable in one equation and substitute the answer in the other equation.

450kg= x + y
Subtract y from both sides
450-y =x


STEP 2:
Substitute (450-y) in second equation in place of x to solve for y.

3y=190kg + x
3y=190 + (450-y)
3y=640 -y
add y to both sides

4y=640
divide both sides by 4

y=160kg Llama weight


STEP 3:
Substitute 160kg in either equation to solve for x.

3y=190kg + x
3(160)=190 + x
480=190 + x
Subtract both sides by 190

290= x
x= 290kg Okapi weight


CHECK:
3y=190kg + x
3(160)=190 + 290
480=480

Hope this helps! :)
8 0
2 years ago
Will works 7 hours a day, 6 days a week.
exis [7]
A day
= £9.20x7
= £64.40

6 days
£64.40x6= £386.40

The wages he has after he shared it with his mom
£386.40/7x5
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= £276

£1932/£276= 7

It will take him 7 weeks to afford a car worth £1932.
6 0
2 years ago
a previous analysis of paper boxes showed that the standard deviation of their lengths is 15 millimeters. A packers wishes to fi
vladimir1956 [14]

Answer:

864.36 boxes

Step-by-step explanation:

In the question above, we are given the following values,

Confidence interval 95%

Since we know the confidence interval, we can find the score.

Z score = 1.96

σ , Standards deviation = 15mm

Margin of error = 1 mm

The formula to use to solve the above question is given as:

No of boxes =[ (z score × standard deviation)/ margin of error]²

No of boxes = [(1.96 × 15)/1]²

= 864.36 boxes

Based on the options above, we can round it up to 97 boxes.

8 0
2 years ago
Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478]. Us
grigory [225]

Answer:

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

\bar X represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n represent the sample size  

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma)

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

5 0
2 years ago
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