S=6a²
a=8mm
S = 6 * 8² = 6 * 64 = 384 mm²
Luke: 1 practice 1 lap, 2 practices 2 laps, 3 practices 3 laps, 4 practices 4 laps
Alex: 1 practice 3 laps, 2 practices 6 laps, 3 practices 9 laps, 4 practices 12 laps
Luke ran 3 laps for 3 practices, and Alex ran 9 laps for 3 practices.
Answer:
The answer is shown in the pic.
Step-by-step explanation:
The cost model will have a fixed cost of $26.30 for 0 ≤ x ≤ 1 lb and variable cost of $4.00/lb for x > 1.
As seen on the pic, the graph will be that of the greatest integer function that is vertically stretched by a factor 4.00 & shifted 26.30 units upward.
Answer:
Step-by-step explanation:
Hello!
X: Cholesterol level of a woman aged 30-39. (mg/dl)
This variable has an approximately normal distribution with mean μ= 190.14 mg/dl
1. You need to find the corresponding Z-value that corresponds to the top 9.3% of the distribution, i.e. is the value of the standard normal distribution that has above it 0.093 of the distribution and below it is 0.907, symbolically:
P(Z≥z₀)= 0.093
-*or*-
P(Z≤z₀)= 0.907
Since the Z-table shows accumulative probabilities P(Z<Z₁₋α) I'll work with the second expression:
P(Z≤z₀)= 0.907
Now all you have to do is look for the given probability in the body of the table and reach the margins to obtain the corresponding Z value. The first column gives you the integer and first decimal value and the first row gives you the second decimal value:
z₀= 1.323
2.
Using the Z value from 1., the mean Cholesterol level (μ= 190.14 mg/dl) and the Medical guideline that indicates that 9.3% of the women have levels above 240 mg/dl you can clear the standard deviation of the distribution from the Z-formula:
Z= (X- μ)/δ ~N(0;1)
Z= (X- μ)/δ
Z*δ= X- μ
δ=(X- μ)/Z
δ=(240-190.14)/1.323
δ= 37.687 ≅ 37.7 mg/dl
I hope it helps!
