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xenn [34]
3 years ago
9

Solve 1/2 x + 3 < 4x - 7

Mathematics
1 answer:
Elanso [62]3 years ago
3 0

Answer:

x > 20/7

good luck on your test <3

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For the annual rate of change of +70%, find the corresponding growth or Decay Factor.
omeli [17]
Hello there!
1. Okay. So 70% is positive, so it is a corresponding growth. 70% is 0.7 in decimal form. C and D are both eliminated, because that's too large. When something grows by 70%, that is less than double. 0.7 is not right either, because that's more of a decay than a growth. The only answer that makes sense id 1.70, because you are going up, and adding 1 to decimal form brings you to the decimal that when multiplied will bring you to the total. The answer is B: 1.70.

2. So, -75% is negative, so it is a decay factor. A is eliminated, because it makes no sense and B is out, because that represents growth, not decay. You lose 75% of the amount overtime compounded, but you still have 25% that remains. To find that amount, you would subtract that amount from 1 to get the decay. 1 - 0.75 is 0.25. The decay factor is 0.25. The answer is D: 0.25.
4 0
3 years ago
Given that the average rate of change for y = f(x) over the interval [0,3] is −1, the average rate of change over the interval [
Svet_ta [14]

Answer:

Answer is 2

Step-by-step explanation:

We know that average rate of change of a function f(x) in the interval (a,b) is

\frac{1}{b-a} \int\limits^a_b {f(x)} \, dx

Using this we can say that

\int\limits^0_3 {f(x)} \, dx =-1(3)=-3\\\int\limits^2_3 {f(x)} \, dx =5(1)=5\\\\\int\limits^2_6 {f(x)} \, dx =4(5)=20\\

Using properties of integration we have

3 to 6 integral = 20-5 =15

0 to2 integral = -3=5 =-8

Thus integral form 0 to 6 would be = -8+15+5 = 12

Average rate of change form 0 to 6 = \frac{12}{6} =2

Answer is 2

7 0
3 years ago
Write the equation for a parabola with a focus at (0,-5) and a directrix at y=-3
Olenka [21]
-4y -16 = x^2
All done
7 0
4 years ago
HELP ASAP PLS!!!
viva [34]
4p(x-4)-5
That is my answer I hope I’m right have a merry Christmas
5 0
3 years ago
Thirty-three college freshmen were randomly selected for an on-campus survey at their university. The participants' mean GPA was
Ivan

Answer: \pm0.1706

Step-by-step explanation:

Given : Sample size : n= 33

Critical value for significance level of \alpha:0.05 : z_{\alpha/2}= 1.96

Sample mean : \overline{x}=2.5

Standard deviation : \sigma= 0.5

We assume that this is a normal distribution.

Margin of error : E=\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}

i.e. E=\pm (1.96)\dfrac{0.5}{\sqrt{33}}=\pm0.170596102837\approx\pm0.1706

Hence, the  margin of error is \pm0.1706

7 0
3 years ago
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