Answer:
They should leave Carlo's house by 4:20 PM
Step-by-step explanation:
The movie starts at 8:10 pm, they want to arrive 20 minutes before the movie starts
20 minutes before 8:10 pm = 7:50 pm.
Hence, they should arrive by 7:50 pm
Duration of their activities before the movie is calculated as:
They will bowl for one hour and dinner will take 1 hour and 15 minutes. Travel time is 20 minutes to the bowling alley, 45 minutes ta the restaurant, and ten minutes to the theater.
: 1 hour + 1 hour 15 minutes + 20 minutes + 45 minutes + 10 minutes
= 2 hour 15 minutes + 75 minutes
= 2 hour 15 minutes +1 hour 15 minutes
= 3 hours 30 minutes in total.
The time they are meant to leave is calculated as:
7:50 pm - 3 hours 30 minutes
= 4:20pm
Answer:
z≥−3
Step-by-step explanation:
1 Subtract 88 from both sides.
5z≥−7−8
2 Simplify -7-8−7−8 to -15−15.
5z≥−15
3 Divide both sides by 55.
z≥− 15/5
4 Simplify 15/5 to 3
z≥−3
Answer:
(15 - 7) * (3 + 5)
12
29
Step-by-step explanation:
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)
