Answer:
19.6, B
Step-by-step explanation:
To solve, we need to figure out how many gallons of white paint she needs for every gallon of brown paint.
We are given this equation.
5 gallons of brown paint = 7 gallons of white paint
Divide both sides by 5.
1 gallon of brown paint = 7/5 gallons of white paint = 1.4 gallons of white paint
Now, using this, substitute 14 into the left side.
14 gallons of brown paint = 1.4 * 14 gallons of white paint
14 gallons of brown paint = 19.6 gallons of white paint
Thus, the answer is B, 19.6 gallons
Answer:
The fraction of one pound of cheese to be used in each sandwich is 1/8
Step-by-step explanation:
Here, we have a question that requires the method of dealing with the division of fractions.
To divide a fraction by a whole number, we multiply the denominator of the fraction by the whole number as follows;
Quantity of cheese Shane has = 1/2 pound
Number of sandwiches to be made = 4
Amount of cheese per sandwich = 1/2÷4 = 1/8.
Answer:
<h2>
The eleventh term of the sequence is 64</h2>
Step-by-step explanation:
The sequence given is an arithmetic sequence
14, 19, 24, …………., 264
The nth term of an arithmetic sequence is given as;
Tn = a+(n-1)d where;
a is the first term = 14
d is the common difference = 19-14=24-19 = 5
n is the number of terms = 11(since we are to look for the eleventh term of the sequence)
substituting the given values in the formula given;
T11 = 14+(11-1)*5
T11 = 14+10(5)
T11 = 14+50
T11 = 64
The eleventh term of the sequence is 64
Answer:
275 non spectators
Step-by-step explanation:
1. Subtract
42,500 - 31,750 - 10,475 =
275
2. Get answer
275 non-spectators
Step-by-step explanation:
let us give all the quantities in the problem variable names.
x= amount in utility stock
y = amount in electronics stock
c = amount in bond
“The total amount of $200,000 need not be fully invested at any one time.”
becomes
x + y + c ≤ 200, 000,
Also
“The amount invested in the stocks cannot be more than half the total amount invested”
a + b ≤1/2 (total amount invested),
=1/2(x + y + c).
(x+y-c)/2≤0
“The amount invested in the utility stock cannot exceed $40,000”
a ≤ 40, 000
“The amount invested in the bond must be at least $70,000”
c ≥ 70, 000
Putting this all together, our linear optimization problem is:
Maximize z = 1.09x + 1.04y + 1.05c
subject to
x+ y+ c ≤ 200, 000
x/2 +y/2 -c/2 ≤ 0
≤ 40, 000,
c ≥ 70, 000
a ≥ 0, b ≥ 0, c ≥ 0.
