Answer:
x = 2.5
Step-by-step explanation:
2x - 11 = -8x + 14
Add 8x to both sides
10x - 11 = 14
Add 11 to both sides
10x = 25
Divide both sides by 10
x = 2.5
Complete question :
Cheddar Cheese
$3/lb
Swiss Cheese
$5/lb
Keisha is catering a luncheon. She has $30 to spend on a mixture of Cheddar cheese and Swiss cheese. How many pounds of cheese can Keisha get if she buys only Cheddar cheese? Only Swiss cheese? A mixture of both cheeses?
What linear equation in standard form can she use to model the situation?
Answer:
10 lbs of cheddar cheese
6 lbs of Swiss cheese
$3a + $5b = $30
Step-by-step explanation:
Given that :
Cheddar cheese = $3/lb
Swiss cheese = $5/lb
Total amount budgeted for cheese = $30
How many pounds of cheese can Keisha get if she buys only Cheddar cheese?
Pounds of cheedar cheese obtainable with $30
Total budget / cost per pound of cheddar cheese
$30 / 3 = 10 pounds of cheedar cheese
Only Swiss cheese?
Pounds of cheedar cheese obtainable with $30
Total budget / cost per pound of Swiss cheese
$30 / 5 = 6 pounds of Swiss cheese
A mixture of both cheeses?
What linear equation in standard form can she use to model the situation?
Let amount of cheddar cheese she can get = a
Let amount of Swiss cheese she can get = b
Hence,
(Cost per pound of cheddar cheese * number of pounds of cheddar) + (Cost per pound of Swiss cheese * number of pounds of Swiss cheese) = total budgeted amount
(3 * a) + (5 * b) = $30
$3a + $5b = $30
Answer:
Step-by-step explanation:
You cannot solve the problem be cause you dont know how many trees there are.
Let Xi be the random variable representing the number of units the first worker produces in day i.
Define X = X1 + X2 + X3 + X4 + X5 as the random variable representing the number of units the
first worker produces during the entire week. It is easy to prove that X is normally distributed with mean µx = 5·75 = 375 and standard deviation σx = 20√5.
Similarly, define random variables Y1, Y2,...,Y5 representing the number of units produces by
the second worker during each of the five days and define Y = Y1 + Y2 + Y3 + Y4 + Y5. Again, Y is normally distributed with mean µy = 5·65 = 325 and standard deviation σy = 25√5. Of course, we assume that X and Y are independent. The problem asks for P(X > Y ) or in other words for P(X −Y > 0). It is a quite surprising fact that the random variable U = X−Y , the difference between X and Y , is also normally distributed with mean µU = µx−µy = 375−325 = 50 and standard deviation σU, where σ2 U = σ2 x+σ2 y = 400·5+625·5 = 1025·5 = 5125. It follows that σU = √5125. A reference to the above fact can be found online at http://mathworld.wolfram.com/NormalDifferenceDistribution.html.
Now everything reduces to finding P(U > 0) P(U > 0) = P(U −50 √5125 > − 50 √5125)≈ P(Z > −0.69843) ≈ 0.757546 .
-4/5 is the skip of the graph
