Asurvey of 200 customers was conducted at a particular cosmic coffee location. about how many of the customers surveyed spend over $5.25 on a drink? remember that the mean price is $4.25 with a standard deviation of $0.67.

This situation can be represented by a standard normal curve with mean $4.25 and std dev $0.67. We need to find the area under this std norm curve that lies above $5.25. It's appropriate to find the pertinent z-score here, as follows:

$5.25-$4.25

z = = 1.49

$0.67

I used the normalcdf function on my TI calculator to determine this area, as follows:

normalcdf(1.49, 100).

The result was 0.05. Thus, about 5%, or 0.05(200 customers), or 10 customers, spend over $5.25 for a drink at this location.

(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