Question

A shipment of racquetballs with a mean diameter of 60 mm and a standard deviation of 0.9 mm is normally distributed. By how many standard deviations does a racquetball with a diameter of 58.2 mm differ from the mean?

Answer 1

@YanaSidlinskiy @uscrnamc @zachcrat can u help?

Answer 2

im working on it

Answer 3

ok, thankyou !! :) @zachcrat

Answer 4

is it multiple choice?

Answer 5

yes.

Answer 6

what are your choices?

Answer 7

0.9
1
2
3

Answer 8

2, because, the SD is .9, and 5.82 is 1.8 away from 60, 1.8/2 is .9 to repeat, 2 is the right answer

Answer 9

thankyou !!! @zachcrat

Answer 10

np

Answer 11

can u help me with another one? @zachcrat

Answer 12

what is it

Answer 13

Multiply: 5x3(2x4 − x3 + 3)

Answer 14

5*3(2*4-3x+3)?

Answer 15

Ok are those exponents??

Answer 16

idk

Answer 17

no, it's 5x^2(2x^4-x^3+3) @zachcrat
& yes @YanaSidlinskiy

Answer 18

5x^3, pellet not ^2

Answer 19

\[5x^2(2x^4-x^3+3)\]
That??

Answer 20

yes

Answer 21

but 5x^3

Answer 22

ok, gemdas
Groups: 2x^4-X^3+3 first

Answer 23

ok?

Answer 24

2x*2x*2x*2x, have you learned the box method?

Answer 25

\[10x^7-5x^6+15x^3\]

Answer 26

these are the options:
7x7 − 5x6 + 15x3
10x7 − x3 + 3
10x7 − 5x6 + 15x3
10x12 − 5x9 + 15x3
so, it's C right?

Answer 27

yup

Answer 28

It's whatever I posted:)

Answer 29

ok. thank y'all so much :)) @zachcrat @YanaSidlinskiy

Answer 30

np