Answer:
The maximum profit can be attained when 4 bikes are produced each day.
Step-by-step explanation:
Look at the attached picture:
In the table given in the picture, the number of bikes produces varies. We cannot properly compare the profits per day. To be consistent, let us determine the profit per unit of bike produced.The relationship between cost, revenue and profit can be as:
Profit = Revenue - Cost
To find the profit per unit of bike, simply divide the profit with the number of bikes produced (1st column). After you see the results, we can see that the highest profit is $17.5 per unit of bike produced. Therefore, the maximum profit can be attained when 4 bikes are produced each day.
Answer:
C. The mean daily salary is greater than $350 per day
Step-by-step explanation:
The computation is shown below:
Y = a + bX
where,
Y = money made by a random selected
a = $150
b = $50
X = number of loan
Now
E(x) = (1 × 0.05) + (2 × 0.10) + (3 × 0.22) + (4 × 0.30) + (5 × 0.18) + (6 × 0.12) + (7 × 0.03)
= 3.94
Now
E(y) = $150 + ($50 × 3.94)
= $347
hence, the option C is not correct
Ummm let me see so what u have to do is
Answer:
∠ZXY and ZX
XYZ is not found on the diagram, as there is no line going from Z to Y, and YX is not a line segement in the diagram, its a full line.
Answer: On average, the weight of a pet visiting this vet on this day is about 2.4 pounds away from 12.9 pounds.
If the MAD of weights for another day was 1.5, then that day's weights would be less variable than the weights of pets seen on this day.
Step-by-step explanation:
Given: The measures in the table describe the weights of animals that visited a vet on one day, in pounds.
Mean = 12.9
Median= 12.0
Mode = 12.0
Mean Absolute Deviation = 2.4
We know that the mean absolute deviation (MAD) of a data set is the mean distance between each and every data value and the mean.It tells about the variation in a data set.
Therefore, On average, the weight of a pet visiting this vet on this day is about 2.4 pounds away from 12.9 pounds.
Also, If the MAD of weights for another day was 1.5, and since 1.5< 2.4.
Then that day's weights would be less variable than the weights of pets seen on this day.
