Answer:
-7a+11
Step-by-step explanation:
Answer:
Yes we can conclude.
Step-by-step explanation:
The sampling distribution of 
can be approximated as a Normal Distribution only if:
np and nq are both equal to or greater than 10. i.e.
Both of these conditions must be met in order to approximate the sampling distribution of as Normal Distribution.
From the given data:
n = 50
p = 0.80
q = 1 - p = 1 - 0.80 = 0.20
np = 50(0.80) = 40
nq = 50(0.20) = 10
This means the conditions that np and nq must be equal to or greater than 10 is being satisfied. So, we can conclude that the sampling distribution of pˆ is approximately a normal distribution
if there are 300 raffle tickets and one raffle ticket is drawn, the proabability of drawing the next one is 1 in 299 because there is one less ticket from the previous win
She drinks 1/4th of the bottle and then another 1/6th of the bottle. So he drinks 3/12+2/12 of a bottle which is 5/12
Answer:
Step-by-step explanation:
2(6x² - 3) = 11x² - x
2*6x² - 2*3 = 11x² - x
12x² - 6 = 11x² -x
Subtract 11x² from both sides
12x² - 11x² - 6 = -x
x² - 6 = -x
x² + x - 6= 0
Sum =1
Product = -6
Factors = 3 , (-2) { 3*(-2) = -6 & 3 +(-2) = 1}
x² + 3x - 2x - 6 = 0
x(x + 3) - 2(x + 3)= 0
(x +3)(x - 2) = 0
