Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
Answer:
x intercept: (1,0)
y intercept: (0,3)
Step-by-step explanation: Substitute variables for 0 and solve
Answer:
The answer is 43 and 19 over 21
Step-by-step explanation:
21 goes into 922 43 times. This giving you 903. 43 Will then be your Whole number. You the subtract 922 by 903, this giving you 19. 19 is going to be your numerator. Then you always keep the denomonator which is 21. This giving you 43 and 19 over 21
Answer:
12,288
Step-by-step explanation:
aₙ = 3 (-2)ⁿ⁻¹
a₁₃ = 3 (-2)¹³⁻¹
a₁₃ = 12,288