The vertical angles are equal to each other, that is, one of the one-sided corners will be 116°.
The sum of the one-sided angles is 180°, which means: 4x+116 = 180°
4x = 180-116 => 4x = 64°
x = 64/4 = 16°.
The sum of two adjacent angles is 180°. which means, that: 2x-3y = 180-64 = 116°
2x-3y = 116°
-2x+3y = -116
-32+3y = -116
3y = -116-(-32)
3y = -84
y = -84/3 => y = -28.
Answer:
C.
Step-by-step explanation:
The random variable W is Normal, with a mean of 28 ounces and a standard deviation of four ounces.
mean of total = 7+7+7+7 = 28
Variance of total = 4+4+4+4 = 16
SD of total = sqrt(16) = 4
The missing part of the question is highlighted in bold format
The Wall Street Journal reported that the age at first startup for 90% of entrepreneurs was 29 years of age or less and the age at first startup for 10% of entrepreneurs was 30 years of age or more.
Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is the sample proportion of entrepreneurs whose first startup was at 29 years of age or less. If required, round your answers to four decimal places. np = n(1-p) = E(p) = σ(p) = (b) Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more. If required, round your answers to four decimal places.
Answer:
(a)
np = 180
n(1-p) = 20
E(p) = p = 0.9
σ(p) = 0.0212
(b)
np = 20
n(1 - p) = 180
E(p) = p = 0.1
σ(p) = 0.0212
Step-by-step explanation:
From the given information:
Let consider p to be the sample proportion of entrepreneurs whose first startup was at 29 years of age or less
So;
Given that :
p = 90% i.e p = 0.9
sample size n = 200
Then;
np = 200 × 0.9 = 180
n(1-p) = 200 ( 1 - 0.9)
= 200 (0.1)
= 20
Since np and n(1-p) are > 5 ; let assume that the data follows a normal distribution ;
Then:
The expected value of the sampling distribution of p = E(p) = p = 0.9
Variance 



The standard error of σ(p) = 

= 0.0212
(b)
Here ;
p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more
p = 10% i.e p = 0.1
sample size n = 200
Then;
np = 200 × 0.1 = 20
n(1 - p) = 200 (1 - 0.1 ) = 180
Since np and n(1-p) are > 5 ; let assume that the data follows a normal distribution ;
Then:
The Expected value of the sampling distribution of p = E(p) = p = 0.1
Variance 



The standard error of σ(p) = 

= 0.0212
Answer:
Step-by-step explanation:
Next time, please try to make your question more specific.
g(x)= -2x^2+5 is a modification of f(x) = x^2 (a parabola with vertex at (0, 0) and opening up).
If f(x) = x^2 is graphed, we can translate its shape up, down, right or left.
In g(x)= -2x^2+5, this graph has been translated up by 5 units, inverted (so that it now opens down), and vertically stretched by a factor of 2.
There are 2 cups to a pint
so multiply by 1/2