First we solve for y₁
3×5 - 5y₁ = 20
15 - 5y₁ = 20
-5y₁ = 20 - 15
-5y₁ = 5
y₁ = -1
(5, -1)

Now we solve for x₂
3x₂ - 5×2 = 20
3x₂ - 10 = 20
3x₂ = 20 + 10
3x₂ = 30
x₂ = 10
(10, 2)

6 0

3 years ago

Assume the given distributions are normal. Cucumbers grown on a certain farm have weights with a standard deviation of 2 ounces.

forsale [732]

Answer:

13.92

Step-by-step explanation:

We have that the critical z-score associated with 85% to the left is 1.04, we know that by table.

So we have to:

m + z * (sd) = 16

where m is the mean, z is the critical z-scor and sd is the standard deviation, if we replace we are left with:

m + 1.04 * (2) = 16

m = 16 - 1.04 * (2)

m = 13.92

Therefore, the average weight if 85% of cucumbers weigh less than 16 ounces is 13.92

4 0

3 years ago