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Murljashka [212]
4 years ago
10

To get the value of an expression you add

Mathematics
2 answers:
frez [133]4 years ago
6 0
When we substitute a specific valuefor each variable, and then perform the operations, it's called evaluating theexpression. Let's evaluate theexpression 3y + 2y when 5 = y. Click on the steps to see how it's done. Perform the operations, to find the value of the expression.
NNADVOKAT [17]4 years ago
6 0
<span>Algebra Basics - Evaluating expressions - First Glance. We have learned that, in in an algebraic expression, letters can stand for numbers. When we substitute a specific value for each variable, and then perform the operations, it's called evaluating the <span>expression.  For more go to

</span></span>www.math.com/school/subject2/lessons/S2U2L3GL.html
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This table on a package of dog food tells how much to feed a dog, depending on its weight.  Weight of Dog (pounds)153045 Amount
Novay_Z [31]

9514 1404 393

Answer:

  k = 2/15

Step-by-step explanation:

We can solve the given equation for k:

  k = s/p . . . . . . divide the given equation by p on both sides

Using the first values from the table (15 pounds, 2 scoops), we have ...

  k = 2/15

The value of k is 2/15.

3 0
3 years ago
Solve each equation.<br> Show all steps<br> 4) -8(-6-5k)=-232
Lubov Fominskaja [6]

the answer is k=4.6. hope this helps

6 0
3 years ago
I am so confused. Can anyone help me??
natita [175]

Answer:

bro!  im in 7th grade i would try to help but im very stupid sorry for the dissopiontment

Step-by-step explanation:

8 0
3 years ago
WILL GIVE BRAINIEST TO VERY FIRST ANSWER
adell [148]
<span>l 18+7 l / 10 - l -16+26 l /6 
= 25 / 10 - 10/6
= 5/2 - 5/3
= 15/6 - 10/6
= 5/6

answer </span><span>5/6 (3rd choice)</span>
8 0
3 years ago
Read 2 more answers
A sample of 20 joint specimens of a particular type gave a sample mean proportional limit stress of 8.41 mpa and a sample standa
nignag [31]

For this problem, the confidence interval is the one we are looking for. Since the confidence level is not given, we assume that it is 95%.

 

The formula for the confidence interval is: mean ± t (α/2)(n-1) * s √1 + 1/n


Where:

<span>
</span>

α= 5%


α/2 = 2.5%


t 0.025, 19 = 2.093 (check t table)


n = 20


df = n – 1 = 20 – 1 = 19

So plugging in our values:


8.41 ± 2.093 * 0.77 √ 1 + 1/20


= 8.41 ± 2.093 * 0.77 (1.0247)


= 8.41 ± 2.093 * 0.789019


= 8.41 ± 1.65141676


<span>= 6.7586 < x < 10.0614</span>

3 0
3 years ago
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