The factors of 50a³ are 1, 2, 5, 10, 25, 50,
and their products with a, a² and a³ .
The factors of 10a² are 1, 2, 5, 10,
and their products with 'a' and a² .
Their common factors are 1, 2, 5, 10,
and their products with 'a' and a².
Their greatest common factor is 10a² .
(Another way to spot it, easily, is to remember this helpful factoid:
If the smaller number is a factor of the larger number,
then the smaller number is their greatest common factor.
Using the greatest common factor, then . . .
50a³ + 10a² = 10a²(5a + 1) .
Y= -5x -4 because you would do slope intercept form and sub the y and x in for -4 and 0
Answer:
a) 615
b) 715
c) 344
Step-by-step explanation:
According to the Question,
- Given that, A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams
- Since the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.
Z = (x - mean)/standard deviation
Now,
For x = 4171, Z = (4171 - 3311)/860 = 1
- P(Z < 1) using Z table for areas for the standard normal distribution, you will get 0.8413.
Next, multiply that by the sample size of 732.
- Therefore 732(0.8413) = 615.8316, so approximately 615 will weigh less than 4171
- For part b, use the same method except x is now 1591.
Z = (1581 - 3311)/860 = -2
- P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.
- For part c, we now need to get two Z scores, one for 3311 and another for 5031.
Z1 = (3311 - 3311)/860 = 0
Z2 = (5031 - 3311)/860= 2
P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772
approximately 47% fall between 0 and 1 standard deviation, so take 0.47 times 732 ⇒ 732×0.47 = 344.
Answer:
this isnt spanish
Step-by-step explanation:
