PLS help due by 3pm
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Answer:
The probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.
Step-by-step explanation:
Let <em>X</em> = number of individuals in the United States who held multiple jobs.
The probability that an individual holds multiple jobs is, <em>p</em> = 0.13.
The sample of employed individuals selected is of size, <em>n</em> = 225.
An individual holding multiple jobs is independent of the others.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 225 and <em>p</em> = 0.13.
But since the sample size is too large Normal approximation to Binomial can be used to define the distribution of proportion <em>p</em>.
Conditions of Normal approximation to Binomial are:
- np ≥ 10
- n (1 - p) ≥ 10
Check the conditions as follows:
The distribution of the proportion of individuals who hold multiple jobs is,
Compute the probability that less than 7.1% of the individuals in this sample hold multiple jobs as follows:
*Use a <em>z</em>-table.
Thus, the probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.
The picture in the attached figure
we have
arc XV = 68°∠YUV = 36°
we know that
The measure of the external angle is the semidifference of the arcs that it covers.
therefore
∠YUV =[arc VY-arc XV]/2
arc VY=[2*∠YUV]+arc XV----------> [2*36]+68---------> 140°
the answer is
the measure of arc VY is 140°
we have
arc XV = 68°∠YUV = 36°
we know that
The measure of the external angle is the semidifference of the arcs that it covers.
therefore
∠YUV =[arc VY-arc XV]/2
arc VY=[2*∠YUV]+arc XV----------> [2*36]+68---------> 140°
the answer is
the measure of arc VY is 140°
Answer:
Step-by-step explanation:
Use cosine:
We have:
Substitute:
<em>cross multiply</em>
<em>divide both sides by 9063</em>