CPhill

This is an isosceles  triangle

The  height  = the length of  the median  drawn from  A

Using the Pythagorean Theorem  .....Height =  sqrt [ 8^2  - 1^2 ]  =  sqrt 63  =  3sqrt 7

AG  = (2/3)  Height   =  (2/3) * 3 sqrt 7   =   2 sqrt 7

5. Heights for 16-year-old girls are normally distributed with a mean of 64 in. and a standard
deviation of 1.9 in.
Find the z-score associated with the 90th percentile.
Find the height of a 16-year-old girl in the 90th perc

z score associated with  .90  = 1.29

( Height -  64 ) /  1.9   =  1.29 

Height   -64  =   1.29 * 1.9

Height  =   1.29 ( 1.9)  + 64   ≈    66.4  in

4. Height of 10th grade boys is normally distributed with a mean of 66.9 in. and a standard deviation
of 2.5 in.
The area greater than the z-score is the probability that a randomly selected 15-year-old boy
exceeds 71 in.
What is the probability that a randomly selected 10th grade boy exceeds 71 in.?
Use your standard normal table.

[ 71  -  66.9  ] / 2.5  =  1.8  = z score

Table  vaue for this  z score  =  .8810

Probability  =  1  - .8810   =  .119  =   11.9%

3. An experiment compares weight of fish for two different brands fish food. Nacho Average Minos
fish food has a mean fish weight 12 lbs., and a standard deviation of 1.4 lbs. Impossible Minos fish
food has a mean fish weight of 14 lbs. and a standard deviation of 1.2 lb. A fish is measured to be
13.1 lbs. Which brand of fish food, Nacho Average Minos, or Impossible Minos, is more likely to have
been used to feed this fish?

 Nacho Average Minos  =  [ 13.1  - 12 ] / 1.4 =  .785

Impossible Minos   =  [ 13.1  - 14 ]  /  1.2 =  - .75

Impossible Minos , because  the fish measured   falls closer to  the  average

2. Body mass index, or BMI, of 18-year-old females are normally distributed with a mean of 24.6 and
a standard deviation of 2.4. Eighteen-year-old Sasha BMI has a z-score of 1.22.
What is Sasha’s BMI?

[  Sash's BMI   -  24.6 ] /  2.4 =    1.22

Sasha's BMI  -24.6  =   1.22 ( 2.4)

Sasha's BMI =   1.22 (2.4)  + 24.6  ≈  27.53

1. Suppose that the weights of 3500 registered female Great Danes in the United States aredistributed normally with a mean of 125 lb. and a standard deviation of 5.2 lb.
Approximately how many of the Great Danes weigh more than 130.2 lbs.?

(130 - 125)  /   5.2  =   z score  of .96  =   table value of .8315

So

3500  ( 1 -.8315)  ≈  135  weigh more

Angles SAP  and SOP  are equal  because they are inscribed angles intercepting the same arc SP

So

SAP  =    SOP

3 *SOP  - 104  =  SOP

3 SOP   -  SOP   =   104

2 SOP  = 104

SOP  = 104 / 2    =   52°

A , H , M , T

Taken one at a time =   4 words

Taken two at a time =  3 + 2 + 1  =  6 words

Taken three at a time  =  2 + 1  = 3  words

Taken four at a time  = 1  word

So   4 + 6 + 3 + 1 =   14  words

Note that the sum of  arc CE + arc DB    =   360  - 107 -144  =  109°

So  CE + DB  = 109     (1)

And   we  have  that, by a geometry theorem

measure of  angle CAE   =   (1/2) (arc CE   - arc DB) )

21  =  (1/2) (CE  - DB)

42 =   CE - DB      (2)

Add  (1)  and (2)

2CE  =  151

CE =   151  / 2    =    75.5°

Remember  that    Distance  / Rate =   Tme

Let the plane speed in still air =  P   

Time  to  fly to tthe meeting  with the wind =  2000 / (P + 50)

Time to return =   2000  / ( P  - 50)

So

     simplify

2000 / (P + 50)   +  2000 / (P -50)  =   7

2000 (P- 50)  + (2000)(P+50)  =   7  (P + 50)(P -50)

4000P   = 7 (P^2   - 2500)

7P^2  -  4000P  - 17500  =  0

Using the Quadratic Formula

4000 ±  sqrt  [ 4000^2  - 4 * 7 * (-17500)  ]                     4000 + 4060.79

_________________________________        =         ________________   =  P   ≈   576 mph    

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