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aivan3 [116]
3 years ago
10

Which ordered pair is a solution of the equation y=5x

Mathematics
2 answers:
Thepotemich [5.8K]3 years ago
7 0
4 days ago and no one answered good job
MrRa [10]3 years ago
6 0

lol the answers are the ones with the stars (***) next to them

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A rectangle has a length of x+2, a width of 7, and a perimeter of 32. The value of x is ?
beks73 [17]
Perimeter = sum of all sides

32 = 2(x+2) + 2(7)
32 = 2x+4 + 14
32 = 2x+18
14 = 2x
 /2     /2

7 = x
4 0
3 years ago
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A bakery sells hollow chocolate spheres. The larger diameter of each sphere is 4cm. The thickness of the chocolate of each spher
Nadusha1986 [10]

Answer:

3.5 cm

Step-by-step explanation:

4 0
3 years ago
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Need help pls<br>thank you in advance ​
ira [324]

Answer:

<h3><u>Mean</u></h3>

<u />

\textsf{Mean}\:\overline{X}=\sf \dfrac{\textsf{sum of all the data values}}{\textsf{total number of data values}}

\implies \sf Mean\:(Nilo)=\dfrac{5+6+14+15}{4}=\dfrac{40}{4}=10

\implies \sf Mean\:(Lisa)=\dfrac{8+9+11+12}{4}=\dfrac{40}{4}=10

<h3><u>Standard Deviation</u></h3>

\displaystyle \textsf{Standard Deviation }s=\sqrt{\dfrac{\sum X^2-\dfrac{(\sum X)^2}{n}}{n-1}}

\begin{aligned}\displaystyle \textsf{Standard Deviation (Nilo)} & =\sqrt{\dfrac{(5^2+6^2+14^2+15^2)-\dfrac{(5+6+14+15)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{482-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{82}{3}}\\\\& = 5.23\end{aligned}

\begin{aligned}\displaystyle \textsf{Standard Deviation (Lisa)} & =\sqrt{\dfrac{(8^2+9^2+11^2+12^2)-\dfrac{(8+9+11+12)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{410-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{10}{3}}\\\\& = 1.83\end{aligned}

<h3><u>Summary</u></h3>

Nilo has a mean score of 10 and a standard deviation of 5.23.

Lisa has a mean score of 10 and a standard deviation of 1.83.

The <u>mean</u> scores are the <u>same</u>.

Nilo's standard deviation is higher than Lisa's.  Therefore, Nilo's test scores are more <u>spread out</u> that Lisa's, which means Lisa's test scores are more <u>consistent</u>.

5 0
1 year ago
World war 1 lasted from August 4th 1914 until November 11th, 1918. During this time an estimated 10 million lives were lost. Abo
Kryger [21]
The question is so dry, mechanical, and devoid of emotion
that it's terrifying.

There is no way to assign a number to "How many people were
dying per day", and I would prefer not even to think about it in
those terms.

-- The period of time from August 4, 1914 until November 11, 1918 is 1,560 days.

-- The "average", or better, the "unit rate" of 10 million events in 1,560 days
is the quotient

                   (10,000,000 events) / (1,560 days) 

               =  6,410.3 events per day

               =     267.1 events per hour

               =       4.45 events per minute.


Reciprocally, this is a unit rate of 

                  13.48 seconds per event,
                  sustained continuously for 4.274 years !

When will we ever learn ! ?
6 0
3 years ago
Can you awnser this for me please
Thepotemich [5.8K]
-2 (Rise over run) goes up 2 and over 1 and the line is going down
3 0
3 years ago
Read 2 more answers
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