Answer:
The z-score for an income of $2,100 is 1.
Step-by-step explanation:
If X 
N (µ, σ²), then 
, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z N (0, 1).
The distribution of these z-variate is known as the standard normal distribution.
Given:
µ = $2,000
σ = $100
<em>x</em> = $2,100
Compute the <em>z</em>-score for the raw score <em>x</em> = 2100 as follows:
Thus, the z-score for an income of $2,100 is 1.
Treat??
hope this helps lol
Notice that half of the coefficient of x is (1/2)(-4), or -2.
Then x^2 - 4x becomes
x^2 - 4x + 4 - 4 = 0, or (x -2)^2 = 4. Then x = -2 plus or minus 2, and the two roots are 0 and -4.
Thus, add 4 to both sides to obtain x^2 - 4x + 4 = 4.
If a,b,c are the 3 positive integers
1/a +1/b +1/c > 6/abc
(bc+ac+ab)/abc >6/abc so
(bc+ac+ab)>6
The lowest positive integers that are different are 1,2,3 so the lowest value that (bc+ac+ab) could have is 1•2+2•3+1•3=2+6+3= 11 therefore
1/a +1/b +1/c > 6/abc is true
Let us make a list of all the details we have
We are given
The cost of each solid chocolate truffle = s
The cost of each cream centre chocolate truffle = c
The cos to each chocolate truffle with nuts = n
The first type of sweet box that contains 5 each of the three types of chocolate truffle costs $41.25
That is 5s+5c+5n = 41.25 (cost of each type of truffle multiplied by their respective costs and all added together)
The second type of sweet box that contains 10 solid chocolate trufles, 5 cream centre truffles and 10 chocolate truffles with nuts cost $68.75
That is 10s+5c+10n = $68.75
The third type of sweet box that contains 24 truffles evenly divided that is 12 each of solid chocolate truffle and chocolate truffle with nuts cost $66.00
That is 12s+12n=$66.00
Hence option C is the right set of equations that will help us solve the values of each chocolate truffle.
