Answer:
0.0035289
Step-by-step explanation:
From the question;
mean annual salary = $63,500
n = sample size = 31
Standard deviation = $6,200
Firstly, we calculate the z-score of $60,500
Mathematically;
z-score = x-mean/SD/√n = (60500-63500)/6200/√(31) = -2.6941
So we want to find the probability that P(z < -2.6941)
We can get this from the standard normal table
P( z < -2.6941) = 0.0035289
The exact value of cos 45 as per the special triangle would be
Cos 45 = 1/✔️2
Rationalizing the denominator gives
✔️2/2 = 0.707.
A) 125 * 10 = 1250 chinchillas in a year;
1250 * 2 = 2500 chinchillas in two years;
b) y = x + 433, where 433 = 933 - 500;
c) 933 + 433 = 1366 chinchillas they have <span>at the end of two years;</span>
Step-by-step explanation:
two people I hope I could help
