Answer:
<em>Cathy was born in 1980 and she was 18 years old in 1998</em>
Step-by-step explanation:
<u>Equations</u>
This is a special type of equations where all the unknowns must be integers and limited to a range [0,9] because they are the digits of a number.
Let's say Cathy was born in the year x formed by the ordered digits abcd. A number expressed by its digits can be calculated as
In 1998, Cathy's age was
And it must be equal to the sum of the four digits
Rearranging
We are sure a=1, b=9 because Cathy's age is limited to having been born in the same century and millennium. Thus
Operating
If now we try some values for c we notice there is only one possible valid combination, since c and d must be integers in the range [0,9]
c=8, d=0
Thus, Cathy was born in 1980 and she was 18 years old in 1998. Note that 1+9+8+0=18
As an estimate the sum may read 720-210, which makes 510
Solution:
Number of times a die is rolled = 20
1 - 3=A
2 - 5=B
3 - 4=C
4 - 2=D
5 - 3=E
6 - 3=F
Total number of arrangements of outcomes , when a dice is rolled 20 times given that 1 appear 3 times, 2 appears 5 times, 3 appear 4 times, 4 appear 2 times , 5 appear three times, and 6 appear 3 times
= Arrangement of 6 numbers (A,B,C,D,E,F) in 6! ways and then arranging outcomes
= 6! × [ 3! × 5! × 4!×2!×3!×3!]
= 720 × 6×120×24×72→→[Keep in Mind →n!= n (n-1)(n-2)(n-3)........1]
= 895795200 Ways
Solution:
we are given that
A student has some $1 and $5 bills in his wallet.
Let there x $1 bills and y number of $5 bills.
He has a total of 14 bills that are worth $34.
So we can write
Now solve these two equations together using substitution as follows
Hence there are 9 $1 bills and 5 bills are of $5.
Answer:
42.22% probability that the weight is between 31 and 35 pounds
Step-by-step explanation:
Problems of normally distributed samples are 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.
In this problem, we have that:
What is the probability that the weight is between 31 and 35 pounds
This is the pvalue of Z when X = 35 subtracted by the pvalue of Z when X = 31. So
X = 35
has a pvalue of 0.5557
X = 31
has a pvalue of 0.1335
0.5557 - 0.1335 = 0.4222
42.22% probability that the weight is between 31 and 35 pounds
