Answer:
We conclude that the mean nicotine content is less than 31.7 milligrams for this brand of cigarette.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 31.7 milligrams
Sample mean,
= 28.5 milligrams
Sample size, n = 9
Alpha, α = 0.05
Sample standard deviation, s = 2.8 milligrams
First, we design the null and the alternate hypothesis

We use One-tailed t test to perform this hypothesis.
Formula:

Putting all the values, we have

Now,
Since,
We fail to accept the null hypothesis and accept the alternate hypothesis. We conclude that the mean nicotine content is less than 31.7 milligrams for this brand of cigarette.
X² + 3x = 10
Convert to standard form.
x² + 3x - 10 = 0
factor x² = x * x
factor -10 = 5 * -2
(x + 5) (x - 2) = 0
x(x -2) + 5(x-2) = 0
x² - 2x + 5x - 10 = 0
x² + 3x - 10 = 0
x + 5 = 0 x² + 3x = 10
x = -5 -5² + 3(-5) = 10
25 - 15 = 10
10 = 10
x - 2 = 0 x² + 3x = 10
x = 2 2² + 3(2) = 10
4 + 6 = 10
10 = 10
Commutative property states that order does not matter. Multiplication and addition are commutative. Related Links: Properties. Associative, Distributive and commutative properties.
Answer:
90m+145
Step-by-step explanation:
80 + 50m + 65 + 40m
90m+145