Answer:
The correct option is C
Step-by-step explanation:
if f(x)= x3 + x2 - 20x
Replace f(x) by y
y = x3 + x2 - 20x
0 =x3 + x2 - 20x
x3 + x2 - 20x = 0
Take out x as a common:
x(x2+x-20)=0
Find factors of x2+x+20.
x(x^2+4x-5x-20) = 0
x{x(x+4)-5(x+4)}=0
x(x+4)(x-5)=0
Set x= 0
x=0 , x+4=0 , x-5 =0
x=0, x=0-4 , x=0+5
x=0, x= -4, x=5
x=(0,5,-4)
The correct option is C....
Answer:
e. The probability of observing a sample mean of 5.11 or less, or of 5.29 or more, is 0.018 if the true mean is 5.2.
Step-by-step explanation:
We have a two-tailed one sample t-test.
The null hypothesis claims that the pH is not significantly different from 5.2.
The alternative hypothesis is that the mean pH is significantly different from 5.2.
The sample mean pH is 5.11, with a sample size of n=50.
The P-value of the test is 0.018.
This P-value corresponds to the probability of observing a sample mean of 5.11 or less, given that the population is defined by the null hypothesis (mean=5.2).
As this test is two-tailed, it also includes the probability of the other tail. That is the probability of observing a sample with mean 5.29 or more (0.09 or more from the population mean).
Then, we can say that, if the true mean is 5.2, there is a probability P=0.018 of observing a sample of size n=50 with a sample mean with a difference bigger than 0.09 from the population mean of the null hypothesis (5.11 or less or 5.29 or more).
The right answer is e.
Answer:
32
Step-by-step explanation:
Step 1: Define
f(x) = 3x² - 5x - 4
g(x) = -4x - 12
Step 2: Find f(g(x))
f(g(x)) = 3(-4x - 12)² - 5(-4x - 12) - 4
f(g(x)) = 3(16x² + 96x + 144) + 20x + 60 + 4
f(g(x)) = 48x² + 288x + 432 + 20x + 64
f(g(x)) = 48x² + 308x + 496
Step 3: Find f(g(-4))
f(g(-4)) = 48(-4)² + 308(-4) + 496
f(g(-4)) = 48(16) - 1232 + 496
f(g(-4)) = 768 - 736
f(g(-4)) = 32
Answer: 3.42 miles
Step-by-step explanation:
5.5 cm converts to 3.41754e-5 miles
Answer: 
<u>Step-by-step explanation:</u>
NOTE: The sign between 4x and 7 is missing in the question you posted. I assumed it was 4x + 7 in order to solve the equation.
