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2 answers:
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Answer:
The probability that the mean of this sample is less than 16.1 ounces of beverage is 0.0537.
Step-by-step explanation:
We are given that the average amount of a beverage in randomly selected 16-ounce beverage can is 16.18 ounces with a standard deviation of 0.4 ounces.
A random sample of sixty-five 16-ounce beverage cans are selected
Let = <u><em>sample mean amount of a beverage</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = population mean amount of a beverage = 16.18 ounces
= standard deviation = 0.4 ounces
n = sample of 16-ounce beverage cans = 65
Now, the probability that the mean of this sample is less than 16.1 ounces of beverage is given by = P( < 16.1 ounces)
P( < 16.1 ounces) = P(
<
) = P(Z < -1.61) = 1 - P(Z
1.61)
= 1 - 0.9463 = <u>0.0537</u>
The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9591.
Answer:
(c) x = 2 and x = -8
Step-by-step explanation:
The rules of logarithms let you rewrite this as a quadratic equation. That equation will have two (2) potential solutions. We know from the domain of the log function that any negative value of x will be an extraneous solution.
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The rules of logarithms that apply are ...
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<h3>take antilogs</h3>We can rewrite the equation so that only one logarithm is involved. Then we can take antilogs.
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<h3>solve the quadratic</h3>Adding (6/2)² = 9 to both sides will "complete the square."
16 +9 = x² +6x +9 . . . . . . . add 9
25 = (x +3)²
±√25 = x +3 = ±5 . . . . . take the square root(s)
x = -3 ±5 = {-8, +2}
The two potential solutions are x = 2 and x = -8.