0.15 * 100 = 15
Answer: 15 times in 100
Answer:
idont know what's the answer but goodluck bro<em><u> </u></em><em><u>loveyou</u></em>
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)
C. \: x = 2 \: or \: x = 3
Answer:
1) A. y = x + 2
Once a line of best fit has been placed upon a scatter graph it is straightforward to find the equation. The general equation of a straight line is: y = mx + c Where m is the slope (gradient) of the line and c is the y-intercept.
4) d. strong positive correlation.
The slope of the line is positive. Most points lie close to the line.
(I am not sure if this is correct or not, I am so sorry)
