Answer:
The store should use 112.5 pounds of Brazilian coffee and 37.5 pounds of Colombian cofee.
Step-by-step explanation:
Let "b" be the amount of Brazilian coffee, in pounds, required for the blend and "c" the amount of Colombian coffee required, in pounds.
Since there are two unknown variables a two-equation system is needed to solve the problem, we can set up one equation for weight and another for price as follows:
Solve for "c" by multiplying the first equation by -10 and adding it to the second one:
Now, solve for b by replacing the value obtained into the first equation
The store should use 112.5 pounds of Brazilian coffee and 37.5 pounds of Colombian cofee.
Answer:
They both have q+3/2p, so that means that 2PQ=CB and that means they are parallel to each other
Step-by-step explanation:
PQ=PA+QA
PQ=1/2(2q-p)+2/5*5p=q-1/2p+2p=q+3/2p
CB=2q+3p=2(q+3/2p)
A1.) <span>the interval about the mean within which 90% of the data lie = 120 + or - 1.645(10) = 120 + or - 16.45 = 120 - 16.45 to 120 + 16.45 = 103.55 to 136.45
2.) P(100 < X < 140) = P(X < 140) - P(X < 100) = P(z < (140 - 120)/10) - P(z < (100 - 120)/10) = P(z < 2) - P(z < -2) = P(z < 2) - [1 - P(z < 2)] = 2P(z < 2) - 1 = 2(0.97725) - 1 = 1.9545 - 1 = 0.9545 = 95.5%
B1.) </span><span>the interval about the sample mean that has a 1% level of confidence is 500 + or - 2.575(80/√1000) = 500 + or - 7 = 500 - 7 to 500 + 7 = 493 to 507
2.) 2P(z < a) - 1 = 0.90
2P(z < a) = 1.90
P(z < a) = 0.95
a = 1.645
(b - 500)/(80/√1000) = 1.645
b - 500 = 4
b = 500 + 4 = 504
The </span><span>interval about the sample mean such that the probability is 0.90 that the mean number lies within the interval is 496 - 504.</span><span />
(3 x 1 ) + (4 x 0.1) + (9 x 0.01) + ( 9 x 0.001)
OR
3 + 0.4 + 0.09 + 0.009
Three and four hundred ninety-nine thousandths
Hope this helped!
Supplementary angles have a degree measure of 180, so they look like a straight line.
Your answers are 4 and 3, 4 and 5, and 3 and 6.
