It’s 12.6, the top and bottom sides are parallel to each other.
Answer:
$11917.03
Step-by-step explanation:
Use Compound Interest Formula:
A = 7,000 (1 + 0.06/2)^2(9)
Simplify:
A = 7,000(1.03)^18
Solve:
A = $11917.03
Answer:
Where Y(t) represent the average annual per capita health care costs
475 represent the initial amount for the average annual per capita health care costs
t represent the number of years since 1970
And 1.076 represent the growth factor given by:
And solving for r we got:
So for this case we can say that the value 1.076 represent the growth factor.
Step-by-step explanation:
For this case we have the following model given:
Where Y(t) represent the average annual per capita health care costs
475 represent the initial amount for the average annual per capita health care costs
t represent the number of years since 1970
And 1.076 represent the growth factor given by:
And solving for r we got:
So for this case we can say that the value 1.076 represent the growth factor.
<h2>
Good morning,</h2>
<em><u>__________________</u></em>
<em><u>Answer</u></em>:
a+a+a = 3a.
4b-b = 3b.
3x²+x² = 4x².
<u><em>explanation</em></u>
a+a+a = 3a.
4b-b = b(4 - 1) = b × 3 = 3b.
3x²+x² = x²(3 + 1) = x² × 4 = 4x².
_______________________________
:)
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.
