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Evgesh-ka [11]
3 years ago
6

Read the excerpt from A Short Walk Around the Pyramids and through the World of Art.

Mathematics
2 answers:
FrozenT [24]3 years ago
8 0

Answer:

b

Step-by-step explanation:

Leokris [45]3 years ago
4 0

<em>Answer:</em>

B. a picture of Knife Edge Mirror Two Piece

<em>Step-by-step explanation:</em>

pls give brainliest, u don't have to

Did it on edge, hope u have a good day :)

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ASAP! Question 3: The following data give the weights (in pounds) lost by 15 members of a health club at the end of 2 months aft
Klio2033 [76]

Answer:

First arrange the data in an order

3, 3, 3, 7, 7, 10, 10, 12, 12, 13, 14, 17, 17, 18, 21

a.) Q1 = (N + 1)/4th item = (15 + 1)/4 = 4th item = 7

Q2 = 2(16)/4 = 16/2 = 8th item = 12

Q3 = 3(16)/4 = 3(4) = 12th item = 17

IQR = (Q3 - Q1)/2 = (17 - 7)/2 = 10/2 = 5

b.) 82nd percentile = 82(16)/100 = 13th item = 17

c.) 12 is the 8th item which is in the middle of the data set, so the percentile rank is 50%

Step-by-step explanation:

3 0
3 years ago
Convert 84% to an equivalent decimal​
dangina [55]

Answer:

To convert a percent to a decmil just multiply the percent by 0.01

84*0.01=0.84

0.84

Hope This Helps!!!

8 0
3 years ago
Read 2 more answers
The 5 in 6.052 has 1/10 the value of the 5 in _____? a)0.815 b)53.2 c)4.45 d)7.5
olasank [31]
Would it be a correct me if im wrong
3 0
3 years ago
In the diagram, LaTeX: \Delta ABC\cong\Delta DFE
leonid [27]

Answer:

x = 13

y = 8

Step-by-step explanation:

8 0
3 years ago
A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with
Ratling [72]

Answer:

P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)

And we can find this probability using the complement rule and with excel or the normal standard table:

P(z>1.61)=1-P(z

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(10.5,0.3)  

Where \mu=10.5 and \sigma=0.3

We are interested on this probability

P(X>10.983)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)

And we can find this probability using the complement rule and with excel or the normal standard table:

P(z>1.61)=1-P(z

8 0
3 years ago
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