Answer:51273.816238
Step-by-step explanation:
Answer:
13.567%
Step-by-step explanation:
We solve the above question, using z score formula
z = (x-μ)/σ, where
x is the raw score = 23.1 ounces
μ is the population mean = 22.0 ounces
σ is the population standard deviation = 1.0 ounce
More than = Greater than with the sign = >
Hence, for x > 23.1 ounces
z = 23.1 - 22.0/1.0
= 1.1
Probability value from Z-Table:
P(x<23.1) = 0.86433
P(x>23.1) = 1 - P(x<23.1)
P(x>23.1) = 1 - 0.86433
P(x>23.1) = 0.13567
Converting to percentage
= 0.13567 × 100
= 13.567%
Therefore, the percentage of regulation basketballs that weigh more than 23.1 ounces is 13.567%
<span>Brenda would be correct, another way to find the answer would be to use the keep change flip method. Where you keep 5/2 change division to multiplication then flip 4/1.
The equation would then be 5/2 * 4/1 then you just multiply straight across 5*4 = 20 and 2*1 equals 2. You end up with 20/2 which reduces to 10</span>
Since the slope is -2 we know that m=-2. Now use the standard form of y=mx+b and plug in the coordinates (-8,2). We know y=2, x=-8, and m=-2. Therefore we get 2=-2(-8)+b. Now solve for b and you get -14. Therefore, the equation is y=-2x-14.
Hope this helped!
