Answer: He can raise up to 40 goats and 100 llamas.
Step-by-step explanation:
Hi, to answer this question we have to write system of equations with the information given:
The space each goat needs (4) multiplied by the number of goats (x); plus The space each llama needs (10) multiplied by the number of llamas must be less or equal to the acre land available (800)
4x +10y ≤ 800 (acres)
The amount of veterinary care (in $) each goat needs (110) multiplied by the number of goats (x); plus The amount of veterinary care each llama needs (88) multiplied by the number of llamas (y)must be less or equal to the Rancher's budget.(14520)
110x +88y ≤ 14,520 (cost)
Multiplying the first equation by 27.5, and subtracting the second equation to the first one:
110x + 275y ≤22,000
-
110x +88y ≤ 14,520
____________
187y ≤ 7480
y ≤ 7480/187
y ≤ 40
Replacing y in the first equation
4x +10(40) ≤ 800
4x +400 ≤ 800
4x ≤ 800-400
4x ≤ 400
x ≤ 400/4
x ≤ 100
Answer:
Omas can buy 1 pound of walnuts.
Step-by-step explanation:
From the information given, you can say that the total cost would be equal to the result of multiplying the price per pound of granola for the number of pounds of granola plus the result of multiplying the price per pound of walnuts for the number of pounds of walnuts, which would be:
Total cost=2x+6y, where
x is the number of pounds of granola
y is the number of pounds of walnuts
Now, you can replace the values on the formula and solve for y:
12=(2*3)+6y
12=6+6y
6=6y
y=1
According to this, the answer is that Omas can buy 1 pound of walnuts.
Answer:
V≈184.31
Step-by-step explanation:
V=πr2h
3=π·42·11
3≈184.30677
Answer:
12%
Step-by-step explanation:
I = Prt
r = I/Pt
r = 792 / 2200(3)
r = 0.12
Answer:
1,082
Step-by-step explanation:
The sample size n in Simple Random Sampling is given by
where
z = 1.645 is the critical value for a 90% confidence level (*)
s = 2 is the estimated population standard deviation
e = 0.1 mm points is the margin of error
so
rounded up to the nearest integer.
So the manufacturer should test 1,083 parts.
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(*)This is a point z such that the area under the Normal curve N(0,1) outside the interval [-z, z] equals 10% = 0.1
It can be obtained in Excel with
NORMINV(1-0.05,0,1)
and in OpenOffice Calc with
NORMINV(1-0.05;0;1)
