Slope = (2+4)/(1-4) = 6/-3 = -2
passing thru (1.2)
y = mx + b
b = y - mx
b = 2 - (-2)(1)
b = 2 + 2
b = 4
equation
y = -2x + 4
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:
<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>
(0.5603,0.6529)
Step-by-step explanation:
<em>Step(i)</em>:-
<em>Given sample size 'n' =300</em>
Given data random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand.
<em>Given sample proportion </em>
<em> </em>
level of significance = 90% or 0.10
Z₀.₁₀ = 1.645
<em>90% confidence interval for the proportion is determined by</em>


(0.6066 - 0.0463 ,0.6066 + 0.0463)
(0.5603,0.6529)
<u>final answer</u>:-
<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>
(0.5603,0.6529)
C. 12 /3 go Brainly! Have Fun