1.2 + d + 0.4 = 9.7
1.6 + d = 9.7
1.6 (-1.6) + d = 9.7 (-1.6)
d = 8.1
your answer is 8.1
hope this helps
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:
t=2.08 seconds.
Step-by-step explanation:
Well, in this example, H(t)=-0.6cos(2pi/2.5)t+1.5 should be equal to 1.2. If calculated, -0.6cos(0.8pi)t=1.2-1.5 which is equal to -0.6cos(0.8pi)t=-0.3, then cos(0.8pi)t=0.5. The value of cosine in terms of radians when it is equal to 0.5 is pi/3. So, cos(0.8pi)t=cos(pi/3). If simplified, (0.8pi)*t=5pi/3. pi's are cancelled out and t is calculated as 2.08333... If rounded to the nearest hundredth it is 2.08.
Answer:
C AND D
Step-by-step explanation:
it is a rational and real number
