The annual commissions per salesperson employed by a retailer of mobile communication devices are normally distributed, and averaged $40,000, with a standard deviation of $5,000. What percent of the salespersons earn between $32,000 and $42,000?

Therefore 60.06% of the salespersons earn between $32,000 and $42,000.

Step-by-step explanation:

Given that:

the mean (μ) is $40000 and

the standard deviation (σ) = $5000

The Z score (z) is given by the equation:

For $42000

For $32000

From the probability table, we find the probability of the z core:

P($32000 < X < $42000) = P(-1.6 < z < 0.4) = P(z < 0.4) - P(z < -1.6) = 0.6554 - 0.0548 = 0.6006 = 60.06%

Therefore 60.06% of the salespersons earn between $32,000 and $42,000.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

none cuz there are no "following statements" regarding the question

step-by-step explanation: