0. At least that's what my phone says...
Vbox-vspheres
vbox=, I assume we are dealing with l=w=h
so
v=lwh=12^3=1728
vsphere=(4/3)pir^3
r=3
vsphere=(4/3)pi3^3=4pi9=36pi
8 of them so
8 times 36pi=288pi or about 904.7786842338604526772412943845 cubic inches
vbox-sphere=1728-904.7786842338604526772412943845=823.2213157661395473227587056155
space filled by packing beads is about 823.22 cubic inches
beads percent is 823.2213157661395473227587056155/1728 times 100=47.64% filled by beads
Answer:
a. The probability that a customer purchase none of these items is 0.49
b. The probability that a customer purchase exactly 1 of these items would be of 0.28
Step-by-step explanation:
a. In order to calculate the probability that a customer purchase none of these items we would have to make the following:
let A represents suit
B represents shirt
C represents tie
P(A) = 0.22
P(B) = 0.30
P(C) = 0.28
P(A∩B) = 0.11
P(C∩B) = 0.10
P(A∩C) = 0.14
P(A∩B∩C) = 0.06
Therefore, the probability that a customer purchase none of these items we would have to calculate the following:
1 - P(A∪B∪C)
P(A∪B∪C) =P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C)
= 0.22+0.28+0.30-0.11-0.10-0.14+0.06
= 0.51
Hence, 1 - P(A∪B∪C) = 1-0.51 = 0.49
The probability that a customer purchase none of these items is 0.49
b.To calculate the probability that a customer purchase exactly 1 of these items we would have to make the following calculation:
= P(A∪B∪C) - ( P(A∩B) +P(C∩B) +P(A∩C) - 2 P(A ∩ B ∩ C))
=0.51 -0.23 = 0.28
The probability that a customer purchase exactly 1 of these items would be of 0.28
Answer:
0.7486 = 74.86% observations would be less than 5.79
Step-by-step explanation:
I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
The general format of the normal distribution is:
N(mean, standard deviation)
Which means that:
What proportion of observations would be less than 5.79?
This is the pvalue of Z when X = 5.79. So
has a pvalue of 0.7486
0.7486 = 74.86% observations would be less than 5.79
Answer:
1) 11
2) 2
3) 
4) 
5) -2
6) 
Step-by-step explanation:
1) 2
+ 3
- 
=(2 × 2 )+ (3 × 4) - 5
= 4 + 12 - 5
= 11
2) 4
-2
+ 
= (4 × 2) - (2 × 7) + 8
= 8 - 14 +8
= 2
3) 5
- 3
= 5× 
- 3×
+ 4×
+ 2×
= 
=
4) 
= 
=
5) 
= 
= -2
6) 
= 
= 2× - 2× - 
= 
=
Hope the working out is clear and will help you. :)
