Answer:
0.6604
Step-by-step explanation:
Given that a market research firm knows from historical data that telephone surveys have a 36% response rate.
Sample size of random sample = 280
We know for samples randomly drawn of large size sample proportion follows a normal distribution with mean= sample proportin and std error
= 
Substitute p = 0.36 and q = 1-0.36= 0.64
p follows N with mean = 0.36 and std dev = 
Using normal distribution values we can find\
If we add 8 oz of glycol to the 40 oz mixture we have 48 oz of mixture and 4+8 = 12 oz glycol
12 / 48 = 1/4 = 25 % glycol
the answer is 8 oz
0^9 +7x+189yx−3y
o
9
+7x−3y
9
+7x+3y
9
−7x−3y
9
−7x+3y
9
+7x+189yx−3y
2 Collect like terms.
{o}^{9}+(7x+7x-7x-7x+7x)+(-3{y}^{9}+3{y}^{9}-3{y}^{9}+3{y}^{9})+189yx-3y
o
9
+(7x+7x−7x−7x+7x)+(−3y
9
+3y
9
−3y
9
+3y
9
)+189yx−3y
3 Simplify.
{o}^{9}+7x+189yx-3y
o
9
+7x+189yx−3y
8*40 = 320 * 2 = $640 for the 2 workers
640/2000 = 0.32
labor cost is 32% of the revenue
The given equation is 
We need to solve the equation for q.
<u>Value of q:</u>
The value of q can be determined by solving the equation for q.
Thus, subtracting both sides of the equation by r, we get;
Now, dividing both sides of the equation by b, we have;
Simplifying the terms, we get;
Therefore, the value of q is 
Hence, Option B is the correct answer.
