Answer:
0.7262
Step-by-step explanation:
Let random variable = X
Probability of getting a cookie with at least 55 chips:
P(55≤ X ≤78), where X1 = 55 and X2 = 78
Let mean, U = 66 and standard deviation, S.D. = 10
Let Z = X - U/S.D; Z1 = X1 - U/S.D and Z2 = X2 - U/S.D
Z1 = 55 - 66/10 = - 1.1
Z2 = 78 - 66/10 = 1.2
P(X1≤ X ≤ X2) = P(X1 -U/S.D ≤ X - U/S.D ≤ X2 - U/S.D)
P(Z1 ≤ Z ≤ Z2) = P(0 ≤ Z ≤ Z1) + P(0 ≤ Z ≤ Z2)
Hence, (55 ≤ X ≤ 78) = P( - 1.1 ≤ Z ≤ 1.2)
∴ P(-1.1 ≤ Z ≤ 1.2) = P(0 ≤ Z ≤ 1.1) + (0 ≤ Z ≤ 1.2)
= 0.3413 + 0.3849 = 0.7262
The probability of the player missing a shot at least once is 1.00 - 0.70 = 0.3.
Therefore the average number of missed double goals in 2500 trials is given by:
Answer:
(3,-3)
Step-by-step explanation:
First you divide 2 out of the first equation and it turns into 5x+y=12.
Then you write one on top of the other so you can add them. It looks like:
5x + y = 12
-5x + y = -18
Then you add them together and the 5x's cancel each other out and you are left with 2y = -6
So then you divide both sides by 2 and get y = -3
Then you plug -3 into the second equation because it is simpler and get -5x - 3 = -18
So then you add three to both sides and get -5x = -15
And finally you divide both sides by -5 and get x=3
Answer:
(1.5, -1.77)
Step-by-step explanation:
See attached worksheet.
We know that the pet Guinea gained 8 ounces in one month.
The question asks for the weight gained by Guinea in also one month.
Since both the described situation and the question both represent the same weight gained in the same duration, therefore, the required integer will be 8.
Answer: 8 ounces.
