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Ronch [10]
3 years ago
7

Shape 1 is similar to shape 2

Mathematics
1 answer:
melamori03 [73]3 years ago
8 0

Yes it is, and don't worry, I hate school too! Amazing username!

You might be interested in
In how many different ways can 10 be written as a sum of odd numbers
kakasveta [241]
5+5=10
3+1+1=10
1+1+1+1+1=10

Hope this Helps!
7 0
3 years ago
a square flower garden in Samantha's backyard has an area of 100 square feet how much fencing will she need to enclose the flowe
leonid [27]

Answer:

perimeter = 40 feet

Step-by-step explanation:

side^2 = 100

side = 10

perimeter = 10 x 4 =40

7 0
3 years ago
In a randomly selected sample of 1169 men ages 35–44, the mean total cholesterol level was 210 milligrams per deciliter with a s
Aneli [31]

Answer:

The highest total cholesterol level a man in this 35–44 age group can have and be in the lowest 10% is 160.59 milligrams per deciliter.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 210, \sigma = 38.6

Find the highest total cholesterol level a man in this 35–44 age group can have and be in the lowest 10%.

This is the 10th percentile, which is X when Z has a pvalue of 0.1. So X when Z = -1.28.

Z = \frac{X - \mu}{\sigma}

-1.28 = \frac{X - 210}{38.6}

X - 210 = -1.28*38.6

X = 160.59

The highest total cholesterol level a man in this 35–44 age group can have and be in the lowest 10% is 160.59 milligrams per deciliter.

5 0
3 years ago
Two semicircles are attached to the sides of a rectangle as shown.
melomori [17]

Answer:

157\ in^{2}

Step-by-step explanation:

we know that

The area of the figure is equal to the area of rectangle plus the area of two semicircles

<u>The area of rectangle is equal to</u>

A=14*5=70\ in^{2}

<u>The area of the small semicircle is equal to</u>

A=\pi r^{2} /2

r=5/2=2.5\ in -----> radius is half the diameter

substitute

A=(3.14)(2.5^{2})/2=9.8125 in^{2}

<u>The area of the larger semicircle is equal to</u>

A=\pi r^{2} /2

r=14/2=7\ in -----> radius is half the diameter

substitute

A=(3.14)(7^{2})/2=76.93\ in^{2}

The area of the figure is equal to

70+9.8125+76.93=156.7425=157\ in^{2}

8 0
3 years ago
Read 2 more answers
Please help me with this​
irinina [24]

Answer:

I don't know if it's two separate questions so I did two

5 0
2 years ago
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