Answer:
-3
Step-by-step explanation:
Here is a different example.
'm' is the gradient (or slope). We measure the difference between the two x values and two y values to see how each of them changed respectively. We know m = rise/run so we can put in our slope (7) and points to find t.
yhttps://www.google.com/url?sa=i&source=images&cd=&cad=rja&uact=8&ved=2ahUKEwjE0ciR9-XhAhWXfysKHfFwDVUQjRx6BAgBEAU&url=http%3A%2F%2Fpassyworldofmathematics.com%2Fgradient-slope-formula%2F&psig=AOvVaw1_FXiYTOdfwHU9OggjGhLw&ust=1556099125501066
Answer:
(a) 283 days
(b) 248 days
Step-by-step explanation:
The complete question is:
The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. (a) What is the minimum pregnancy length that can be in the top 11% of pregnancy lengths? (b) What is the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths?
Solution:
The random variable <em>X</em> can be defined as the pregnancy length in days.
Then, from the provided information
.
(a)
The minimum pregnancy length that can be in the top 11% of pregnancy lengths implies that:
P (X > x) = 0.11
⇒ P (Z > z) = 0.11
⇒ <em>z</em> = 1.23
Compute the value of <em>x</em> as follows:

Thus, the minimum pregnancy length that can be in the top 11% of pregnancy lengths is 283 days.
(b)
The maximum pregnancy length that can be in the bottom 5% of pregnancy lengths implies that:
P (X < x) = 0.05
⇒ P (Z < z) = 0.05
⇒ <em>z</em> = -1.645
Compute the value of <em>x</em> as follows:

Thus, the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths is 248 days.
Ok so the formula for this is A=1/2bh lets plug it in.
A=1/2*16*8
A=8*8
A=64 in^2
Answer:
k = 11
Step-by-step explanation:
k/90 - 1/9 = 1/90
k/90 = 1/90 + 1/9
k/90 = 1 + 10/90
k/90 = 11/90
k = 11/90 / 90
k = 11/90 * 90
k = 11
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