Answer:
Our answer is 0.8172
Step-by-step explanation:
P(doubles on a single roll of pair of dice) =(6/36) =1/6
therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)
=1-(1-1/6)3 =91/216
for 12 players this follows binomial distribution with parameter n=12 and p=91/216
probability that at least 4 of the players will get “doubles” at least once =P(X>=4)
=1-(P(X<=3)
=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)
=1-0.1828
=0.8172
Answer:
95% z-confidence interval for the proportion of all children enrolled in kindergarten who attended preschool is between a lower limit of 0.528 and an upper limit of 0.772.
Step-by-step explanation:
Confidence interval = p + or - zsqrt[p(1-p) ÷ n]
p is sample proportion = 39/60 = 0.65
n is the number of children sampled = 60
Confidence level (C) = 95% = 0.95
Significance level = 1 - C = 1 - 0.95 = 0.05
Divide significance level by 2 to obtain critical value (z)
0.05/2 = 0.025 = 2.5%
z at 2.5% significance level = 1.96
zsqrt[p(1-p) ÷ n] = 1.96sqrt[0.65(1-0.65) ÷ 60] = 1.96sqrt[0.2275 ÷ 60] = 1.96sqrt(3.792×10^-3) = 1.96×0.062 = 0.122
Lower limit = p - 0.122 = 0.65 - 0.122 = 0.528
Upper limit = p + 0.122 = 0.65 + 0.122 = 0.772
95% confidence interval is between 0.528 and 0.772
Answer:
f(x)= -(x+1)^2 -2
Step-by-step explanation:
f(x)=(x-h)^2 +k; (h,k)
Then you would substitute the x and y of the coordinate and solve.
-6= -(1+1)^2 -2
-6= -(4) -2
-6 = -6
Answer:
h=48/x
Step-by-step explanation:
