Answer:
The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
For this problem, we have that:

93% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).
f(x) = x³ - 2x² - 24x
x³ - 2x² - 24x = 0
x(x² - 2x - 24) = 0
x = 0
x² - 2x - 24 = 0
a = 1, b = -2, c = -24
Delta = (-2)² - 4 * 1 * (-24) = 4 + 96 = 100
x = (-(-2) - 10)/(2 * 1) = -8/2 = -4
x = (-(-2) + 10)/(2 * 1) = 12/2 = 6
Answer: 2)
I agree with the second answer. Was this correct?
Answer:here's ur answer
Step-by-step explanation:
Answer: 4.F(x) > 0 over the intervals (-0.7, 0.76) and (0.76, ∞).
Hope this helps :)