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4 years ago

Write a situation that can be represented by the expression 3a+6 where a=age in years

1 answer:

7 0

This equation can be used when comparing ages.

An example to illustrate this:
Assume that adding 6 to 3 times the age of Jack will give us the age of his grandfather.
When translating this into equations, assuming that the age of jack is "a" and the age of his grandfather is "b", we will find that:
b = 6 + 3a

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Help me with this lol value of( 2^3)^2

ss7ja [257]

Answer:

Step-by-step explanation:

first, we'll start with the value inside the parenthesis, which is 2³

2³ means to multiply 2 by itself 3 times

so it would be

2×2×2=8

which then means that the expression is now (8)²

now we need to do (8)², which is to multiply 8 by itself twice

8*8=64

hope this helps!

6 0

3 years ago

I spent 80% of my money and had $12 left. How much money did I have to start ?

xeze [42]

If you spent 80% that means you have 20% left, we'll use B as variable

20% x B = 12
0.20/0.20 x B = 12/0.20
B = 60

6 0

4 years ago

Quincy found 120 shells during four days at the beach. if quincy found the same number of shells each day, how many shells did h

Svetach [21]

If Quincy finds 120 shells in 4 days, we can use the equation 4x = 120 to find how many shells she found each day. 4x = 120 If we divide both sides by four we get x = 30 So Quincy found 30 shells each day.

5 0

4 years ago

Read 2 more answers

An estimated regression equation that was fit to estimate ductility in steel using its carbon content was found to be significan

AnnZ [28]

Answer:

At a level of 95%, it is expected that the interval [0.45; 11.59] contains the value of the ductility in steel when its carbon content is 0.5%.

Step-by-step explanation:

Hello!

Considering the dependent variable:

Y: Ductility in steel.

And the independent variable:

X: Carbon content of the steel.

The linear regression was estimated and a prediction interval was calculated.

The prediction interval is calculated to predict a value that the variable Y (response variable) can take for a given value of the variable X (predictor variable) in the definition range of the linear regression line. Symbolically [Y/X= $x_{0}$ ]

In this case 95% prediction interval for Y/X=0.5

At a level of 95%, it is expected that the interval [0.45; 11.59] contains the value of the ductility in steel when its carbon content is 0.5%.

I hope it helps!

6 0

3 years ago

Your company spends 15% of their budget on rent each year. Their operating budget was $45,000.

ella [17]

They have 6750 dollars left

6 0

3 years ago