We have two equations that were solved by Nikki and Jonathon:
Equating the above two:
⇒ 1.3x + 1.6 = -2.7x + 3.2
⇒ 4x = 1.6
⇒ x = 0.4
Hence, substituting the value of x in one of the equations we get:
y = 1.3×0.4 + 1.6 = 2.12
So the solution is (0.4, 2.12)
Jonathon's solution was (0.4, 2.12) and Nikki's was (2.25, 0.5). Hence Jonathon gave the correct solution.
Yesterday Hadi earned: $120
Today he will earn: $120*85% = $102
(You times the amount yesterday by 85% because you want to find the 15% less of 100% of the amount he's earned yesterday. Therefore, to find 15% less than yesterday you multiply it by 85% (100% - 15%) )
Tomorrow he will earn: $102*115% = $117.30
(Same thing here, however you are finding 15% more than 100% today. Therefore, you multiply the amount he earned today it by 115% (100% + 15%) )
To find the amount he earns for the three days you must add the amount earned of yesterday, today and tomorrow earned from the working above.
Therefore, Hadi expects to earn:
Yesterday + Today + Tomorrow = 120 + 102 + 117.30 = $339.3
Use the ! tool to find the # of combinations.
8!/5! = 40,320/120 = 336
Answer: B) 
C) 
Step-by-step explanation:
Thus, Option B and C is giving the result 4b,
Therefore, they are the expressions that are equivalent to 4b.
Answer:
a)0.099834
b) 0
Step-by-step explanation:
To solve for this question we would be using , z.score formula.
The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.
A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is normal with mean 21.37 and variance 0.16.
a) Find the probability that the weight of a single mint selected at random from the production line is less than 20.857 grams.
Standard Deviation = √variance
= √0.16 = 0.4
Standard deviation = 0.4
Mean = 21.37
x = 20.857
z = (x-μ)/σ
z = 20.857 - 21.37/0.4
z = -1.2825
P-value from Z-Table:
P(x<20.857) = 0.099834
b) During a shift, a sample of 100 mints is selected at random and weighed. Approximate the probability that in the selected sample there are at most 5 mints that weigh less than 20.857 grams.
z score formula used = (x-μ)/σ/√n
x = 20.857
Standard deviation = 0.4
Mean = 21.37
n = 100
z = 20.857 - 21.37/0.4/√100
= 20.857 - 21.37/ 0.4/10
= 20.857 - 21.37/ 0.04
= -12.825
P-value from Z-Table:
P(x<20.857) = 0
c) Find the approximate probability that the sample mean of the 100 mints selected is greater than 21.31 and less than 21.39.
