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Lena [83]
3 years ago
10

Help please if u can!

Mathematics
1 answer:
Tamiku [17]3 years ago
7 0

Answer:

idk

Step-by-step explanation:


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Lowest common multiple of 9
ch4aika [34]
Is 3.

3×3=9

Hope this helps!
3 0
3 years ago
The vertices of ABC are A(-2, 2), B(6,2), and CO, 8). The perimeter of ABC?
Alexxx [7]

Answer:

Perimeter = 22.809836694575                              

Step-by-step explanation:

A(-2, 2)  ;  B(6,2)  ;  C(0, 8)

AB=\sqrt{\left( 6--2\right)^{2}  +\left(2-2 \right)^{2}  } =\sqrt{64} =8

AC=\sqrt{\left( 0--2\right)^{2}  +\left(8-2 \right)^{2}  } =\sqrt{40} =2\sqrt{10}

BC=\sqrt{\left( 0-6\right)^{2}  +\left(8-2 \right)^{2}  } =\sqrt{72} =6\sqrt{2}

Then

The perimeter of ΔABC = AB + BC + AC

                                       =8+6\sqrt{2} +2\sqrt{10}

                                       = 22.809836694575

3 0
2 years ago
Evaluate the following expression when x = 3 and y = 4:
JulijaS [17]
I don't understand.  What are the values given?  Are they values of an x,y plot?
If you plot them they are a straight line with y =6x - 2.9  but I don't know what you mean by when x=3 and y = 4?

5 0
3 years ago
Find the area of a circle having a circumference of 231 pi units. Round to the nearest tenth.
loris [4]
Do you by any chance have a diagram of the circle or the r value?
7 0
3 years ago
All cars can be classified into one of four​ groups: the​ subcompact, the​ compact, the​ midsize, and the​ full-size. There are
Olegator [25]

Answer:

p value = 0.302

Step-by-step explanation:

Given that all cars can be classified into one of four​ groups: the​ subcompact, the​ compact, the​ midsize, and the​ full-size. There are five cars in each group. Head injury data​ (in hic) for the dummies in the​ driver's seat are listed below.

H_0: All cars have same mean values

H_a: atleast two cars have different mean values

(Two tailed anova test)

Anova: Single Factor      

     

SUMMARY      

Groups Count Sum Average Variance  

Subcompact 5 3444 688.8 48502.7  

compact 5 2879 575.8 4582.7  

Midsize 5 2534 506.8 18720.2  

Full size 5 2689 537.8 23905.2  

     

     

ANOVA      

Source of Variation SS df MS            F             P-value F crit

Between Groups 94825 3 31608.33 1.321    0.302           3.24

Within Groups 382843.2 16 23927.7    

     

Total 477668.2 19    

Since p value of 0.302 is greater than 0.05 significance level we accept null hypothesis.

p value = 0.302

6 0
3 years ago
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