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Dimas [21]
3 years ago
15

Consider the following graph of a quadratic function.

Mathematics
1 answer:
allochka39001 [22]3 years ago
6 0

Answer:( x-4) ^2+3

Step-by-step explanation:Shifted to the right by 4. Shifted upward by 3

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What is the greatest<br> common factor (GCF)<br> of 18 and 54?
Mariana [72]

Answer:

GCF of 18 and 54 is 18.

Step-by-step explanation:

5 0
2 years ago
Please helppp with this math question &lt;333
Naya [18.7K]

Answer:

The average rate of change of the function g(x)=x^2+10x+18 over the interval -11 \leq x\leq -1 is -1

Step-by-step explanation:

We are given the function g(x)=x^2+10x+18 over the interval -11 \leq x\leq -1

We need to find average rate of change.

The formula used to find average rate of change is : Average \ rate \ of \ change=\frac{g(b)-g(a)}{b-a}

We have b=-1 and a=-11

Finding g(b) = g(-1)

g(x)=x^2+10x+18\\Putting \ x=-1\\g(-1)=(-1)^2+10(-1)+18\\g(-1)=1-10+18\\g(-1)=9

Finding g(a) = g(-11)

g(x)=x^2+10x+18\\Putting \ x=-11\\g(-11)=(-11)^2+10(-11)+18\\g(-1)=121-110+18\\g(-1)=29

Finding average rate of change

Average \ rate \ of \ change=\frac{g(b)-g(a)}{b-a}\\Average \ rate \ of \ change=\frac{9-29}{-1-(-11)}\\Average \ rate \ of \ change=\frac{-10}{-1+11}\\Average \ rate \ of \ change=\frac{-10}{10}\\Average \ rate \ of \ change=-1

So, the average rate of change of the function g(x)=x^2+10x+18 over the interval -11 \leq x\leq -1 is -1

5 0
3 years ago
Richard receives a 3.5% commission for selling cleaning supplies. What is his commission on sales of $122,591?
konstantin123 [22]

Answer:

We have to find 3.5% $122,591

So,

122591*3.5/100=4,290.685

So, the commission will be $4,290.685

5 0
3 years ago
Read 2 more answers
Previously, an organization reported that teenagers spent 24.5 hours per week, on average, on the phone. The organization thinks
Lena [83]

Answer:

We need to develop a one-tail t-student test ( test to the right )

We reject H₀  we find evidence that student spent more than 24,5 hours on the phone

Step-by-step explanation:

Sample size  n = 15     n < 30

And we were asked if the mean is higher than, therefore is a one-tail t-student test ( test to the right )

Population mean   μ₀  = 24,5

Sample mean   μ  =  25,7

Sample standard deviation s = 2

Hypothesis Test:

Null Hypothesis      H₀                             μ  =  μ₀

Alternative Hypothesis     Hₐ                  μ  >  μ₀

t (c) =  ?

We will define CI = 95 %  then   α = 5 %   α = 0,05    α/2 =  0,025

n = 15     then degree of freedom    df = 14

From t-student table  we get:  t(c) = 2,1448

And  t(s)

t(s) = ( μ  -  μ₀  ) / s/√n

t(s) = (25,7 - 24,5) /2/√15

t(s) = 2,3237

Now we compare   t(c)   and  t(s)

t(c)  =  2,1448         t(s)  = 2,3237

t(s) > t(c)

Then we are in the rejection region we reject H₀   we have evidence at 95% of CI that students spend more than 24,5 hours per week on the phone

8 0
3 years ago
If g(x)=<br> x+1<br> and h(x) = 4 – x, what is the value of (9.0) (-3)?
konstantin123 [22]

Answer:it’s b

Step-by-step explanation:

On edn

6 0
3 years ago
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