Answer: d = -3
Step-by-step explanation:
2d+3=7d+18
-5d=15
d=-3
Answer:
P ( 5 < X < 10 ) = 1
Step-by-step explanation:
Given:-
- Sample size n = 49
- The sample mean u = 8.0 mins
- The sample standard deviation s = 1.3 mins
Find:-
Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.
Solution:-
- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:
X ~ N ( u , s /√n )
Where
s /√n = 1.3 / √49 = 0.2143
- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:
P ( 5 < X < 10 ) = P ( (5 - 8) / 0.2143 < Z < (10-8) / 0.2143 )
= P ( -14.93 < Z < 8.4 )
- Using standard Z-table we have:
P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1
I think that is is 8.3, but I could be incorrect!
Replace x and y with the given values:
2(1/8 + 4) - (3/16*8)
Use distributive property:
1/4 + 8 - (3/16 * 8)
Simplify to get final answer:
8 1/4 - 1 1/2 = 6 3/4
Answer:
210
Step-by-step explanation:
For getting this answer we need to find the minimum common multiplier, this is, the smaller number that is multiple of all 5, 6 and 7.
For example, 30 is multiple of both 5 and 6, son in 30 days two suppliers will come, but not the third, as 30 is not a multiple of 7.
Let's try with the next multiple of 5 and 6, this is 60, but is not a multiple of 7.
The next is 90, but again not multiple of 7. Niether are the next: 120, 150, 180.
But when we arrive to 210 we see that, as we are going from multiples of 5 and 6, and 210/7=30, 210 is also multiple of 7.
So, the answer is 210.
