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Aleonysh [2.5K]
3 years ago
15

If you help me I will help you exhausted

Mathematics
2 answers:
saul85 [17]3 years ago
6 0

i agree and i'm in 6th grade too. and this is very easy to me.....but i fell bad for u.

torisob [31]3 years ago
3 0
5,120 because volume is area of base×height
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Can someone please help me get brainliest?? please?
gogolik [260]

Answer:

sure, why not

Step-by-step explanation:

6 0
3 years ago
John has a rectangular-shaped field whose length is 62.5 yards and width is 45.3 yards. The area of the field is ________ square
oee [108]
Area = 2831.25 square yards

Perimeter =215.6 yards

EXPLANATION

The area and perimeter of a rectangular field are found using the formula for finding the area and perimeter of a rectangle respectively.

That means, area of the rectangular field is given by the formula,

A= l\times w

We just have to substitute
l=62.5 and w= 45.3 into the given formula and evaluate.

This implies that;

A= 62.5\times 45.3

This gives the area of the rectangular-shaped field to be;

A= 2831.25 square yards.

Now for the perimeter, we use the formula

P=2w +2l

Or

P=2(w +l)

Substituting the values for the length and width gives,

P=2(62.5+45.3)

\Rightarrow P=2(107.8)

\Rightarrow P=215.6

Hence the perimeter of the rectangular shaped field is 215.6 yards.
8 0
3 years ago
The league championship 800 meter race has 6 runners.
s2008m [1.1K]

Answer: 120

Step-by-step explanation:

Given : The league championship 800 meter race has 6 runners.

The number of positions = 3

Since order matters here , so we use Permutation.

The permutation of n things taking r at a time is given by :-

^nP_r=\dfrac{n!}{(n-r)!}

Then, the number of different ways they place first, second, and third :_

^6P_3=\dfrac{6!}{(6-3)!}=\dfrac{6\times5\times4\times3!}{3!}=120

Hence, they can place first, second, and third in 120 different ways .

7 0
3 years ago
Please help... i need it
Tanya [424]
Oh ok .so the answer is.....2838384764733 I suggest the method of comparing variables
8 0
3 years ago
Read 2 more answers
In performing a chi-square goodness-of-fit test for a normal distribution, a researcher wants to make sure that all of the expec
Sholpan [36]

Answer:

The degrees of freedom for the chi square statistic is 7-4=3

3 0
3 years ago
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