For a polynomial of the form ax^2+bx+c rewrite the middle term as a sum of two terms whose product is a⋅c=5⋅4=20 and whose sum is b=12.
<u>Factor 12 out of 12x.</u>
5x^2+12(x)+4
<u>Rewrite 12 as 2 plus 10</u>
5x^2+(2+10)x+4
Apply the distributive property.
5x^2+2x+10x+4
Factor out the greatest common factor from each group.
Group the first two terms and the last two terms.
(5x^2+2x)+10x+4
Factor out the greatest common factor (GCF) from each group.
x(5x+2)+2(5x+2)
Factor the polynomial by factoring out the greatest common factor, 5x+25x+2.
(5x+2)(x+2)
The answer would be D) x = 11.
Add 5 to both sides (5x = 50 + 5)
Simplify 50 + 5 to 55 (5x = 55)
Divide both sides by 5 (x = 55/5)
Simplify 55/5 to 11 (x = 11)
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:
A. 0.2, 0.25, 0.2
B. the scores are a quarter of the whole
