Answer: 72
<u>Step-by-step explanation:</u>
There are 9 students to choose from to go the library. After that person is chosen there are 8 students remaining.
First student and Second student
9 x 8 = 72
Answer:
Option C is the correct answer.
Step-by-step explanation:
Looking at the functions given,
Initial amount deposited into the account is $150 This means that the principal is
P = 150
It was compounded quarterly. This means that it was compounded 4 times in a year. So
n = 4
The rate at which the principal was compounded is 3%. So
r = 3/100 = 0.03
It was compounded for x years. So
t = x years
The formula for compound interest is
A = P(1+r/n)^nt
A = total amount in the account at the end of t years. Therefore, the
function that models the value in x years of an investment at 3% annual interest compounded quarterly would be
150 (1+0.03/4)^4×x
150 (1 +.0075)^4x
Answer:
One
General Formulas and Concepts:
<u>Algebra I</u>
Slope-Intercept Form: y = mx + b
- m - slope
- b - y-intercept
Solving systems of equations
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify systems</em>
y = 3x + 5
y = 4x
<u>Step 2: Solve</u>
If we compare the 2 lines, we can see that they both have a different slope. If they had the same slope but different y-intercepts, then they would be parallel and have no solution. We can also see that the 2 lines aren't the same. If they were, then they would have infinite solutions.
∴ the systems should have only one solution.
Answer:
Step-by-step explanation:
The null hypothesis is
H0: μ = 32.6
The alternative hypothesis is
Ha: μ ≠ 32.6
The calculated test statistic is 2.66 for the right tail and - 2.66 for the left tail
Since the critical values for both tails is ± 2.145, we would compare the critical values with the test statistic values
In order to reject the null hypothesis, the test statistic must be smaller than - 2.145 or greater than 2.145
Since - 2.66 < - 2.145 and 2.66 > 2.145, we would reject the null hypothesis.
Therefore, Reject H0. There is sufficient evidence to support the claim that the mean is different from 32.6.
2) Test statistic: t = 2.66. Critical values: t = ±1.96
Since - 2.66 < - 1.96 and 2.66 > 1.96, we would reject the null hypothesis. Then
Reject H0. There is sufficient evidence to support the claim that the mean is different from 32.6.