The candy store owner should use 37.5 pounds of the candy costing $1.25 a pound.
Given:
- Candy costing $1.25 a pound is to be mixed with candy costing $1.45 a pound
- The resulting mixture should be 50 pounds of candy
- The resulting mixture should cost $1.30 a pound
To find: The amount of candy costing $1.25 a pound that should be mixed
Let us assume that the resulting mixture should be made by mixing 'x' pounds of candy costing $1.25 a pound.
Since the total weight of the resulting mixture should be 50 pounds, 'x' pounds of candy costing $1.25 a pound should be mixed with '
' pounds of candy costing $1.45 a pound.
Then, the resulting mixture contains 'x' pounds of candy costing $1.25 a pound and '' pounds of candy costing $1.45 a pound.
Accordingly, the total cost of the resulting mixture is 
However, the resulting mixture should be 50 pounds and should cost $1.30 a pound. Accordingly, the total cost of the resulting mixture is 
Equating the total cost of the resulting mixture obtained in two ways, we get,
This implies that the resulting mixture should be made by mixing 37.5 pounds of candy costing $1.25 a pound.
Learn more about cost of mixtures here:
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Answer:
x = 12
Step-by-step explanation:
In similarity triangles angles are congruent.
∠D = ∠A
x² - 8x = 48
x² - 8x - 48 = 0
x² + 4x - 12x - 48 = 0
x(x + 4) - 12(x + 4) = 0
(x +4)(x - 12) = 0
x - 12 = 0 {Ignore x + 4 = 0, as measurements won't have negative values}
x = 12
Correct Answer:Option A. 0.01
Solution:This is a problem of statistics and uses the concept of normal distributions. We need to convert the score of 90 into z-score and then find the desired probability from standard normal distribution table.
Converting 90 to z-score:
Now we are to find the probability of z score being more than 2.33. From the z-table the probability comes out to be 0.01.
Therefore, we can conclude that the probability of class average is greater than 90 is 0.01.
Answer:
even I think
Step-by-step explanation:
Answer:
Step-by-step explanation:
This is simply a units conversion problem. It gives us for the number of passengers, the number of seats per carriage and the number of carriages per train. To change the units from passengers to trains without changing the value, we use the multiplicative identity (that is, 1).
350000 passengers
(350000 passengers) * 1
(350000 passengers) * ((1 carriage)/(32 passengers)) * ((1 train)/(15 carriages)
[note: passengers and carriages cancel. Leaving only trains]
(350000)*(1/32)*(1/15) trains [note: I write this way to paste into MS Excel]
729.1667 trains [oh, but don’t just round this number either up or down]
729 full trains can carry 729*32*15 = 349920 passengers
730 full trains can carry 730*32*15 = 350400 passengers
Now, we can say that 730 trains are adequate to carry 350000 passengers.
