<span>–1 + 6(–1 – 3x) > –39 – 2x.
</span>-1-6-18x>-39-2x
-7-18x>-39-2x
-18x>-32-2x
-16x>-32
x<2
B. x<2
Answer:
The correct answer is
(0.0128, 0.0532)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence interval
, 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:
In a random sample of 300 circuits, 10 are defective. This means that
and 
Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.
So
= 0.05, 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 correct answer is
(0.0128, 0.0532)
Answer:
I think its D
Step-by-step explanation:
He ate 1 - 6/8 which is 2/8, or 1/4 of an orange.
Answer:
Step-by-step explanation:
Hello!
X: number of absences per tutorial per student over the past 5 years(percentage)
X≈N(μ;σ²)
You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.
The formula for the CI is:
X[bar] ±
* 
⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.
Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4
X[bar]= 10.41
S= 3.71

[10.41±1.645*
]
[9.26; 11.56]
Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.
I hope this helps!