There were 1150 pounds of flour in the grain store. After most of the flour was sold, there were 230 pounds left. What was the p
2 answers:
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Answer:
and
Step-by-step explanation:
Given:
Equation 1:
Simplifying the above equation by dividing both sides by 100
Equation 2:
Simplifying the above equation by dividing both sides by 400.
Now the system of equation is:
(1a)
(2a)
Solving by elimination
Multiplying equation (2a) with
(2b)
Adding equations (1a) and (2b) in order to eliminate
+
We get
∴
Plugging in equation (2a).
Adding both sides.
∴
Answer:
Number of units of first product = 112
Number of units of second product = 400
Step-by-step explanation:
Given that there are two products which are to be shipped.
<em>First product details:</em>
Unit volume = 10 cu ft
Weight of one unit = 50 lb
<em>Second product details:</em>
Unit volume = 3 cu ft
Weight of one unit = 40 lb
Total volume that can be carried by a truck = 2320 cu ft
Total weight that can be carried by a truck = 21600 cu ft
To find:
The number of units of each product that can be transported in one shipment in the truck?
Solution:
Let number of units of first product =
Let number of units of second product =
We can write down two linear equations in two variables here and solving them will give us the answer we are required to find.
As per question statement, the equations can be written as:
Multiplying equation (2) with 2 and subtracting equation (1) from it:
By equation (1):
Answer is:
Number of units of first product = 112
Number of units of second product = 400
Answer:
The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.
Step-by-step explanation:
Use the trick of transforming the given random variable one that is standard normal distributed, aka a z-score, then look up the two percentile values in z-score tables to get two equations with two unknowns.
So, let x be the variable describing the marks of a student, and
the standardized equivalent of that (mu - mean, sigma - standard deviation).
We are looking for values of mu and sigma. At this point we'd be out of luck, but, wait, we're given two bits of info: the 10% point (aka, 90th percentile) and the 20% point (can be interpreted as 100-20th percentile). For each point we can use z-score tables to look up the corresponding values of z (just search for z tables). I found:
z-score for the 10% point: z_10=1.28
z-score for the 20% point: z_20=-0.84
That gives us two equations:
and can be solved for mu and sigma (do the work on your end, I am showing my result):
The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.