a. Find the probability that an individual distance is
greater than 214.30 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (214.30 – 205) / 8.3
z = 1.12
Since we are looking for x > 214.30 cm, we use the
right tailed test to find for P at z = 1.12 from the tables:
P = 0.1314
b. Find the probability that the mean for 20 randomly
selected distances is greater than 202.80 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (202.80 – 205) / 8.3
z = -0.265
Since we are looking for x > 202.80 cm, we use the
right tailed test to find for P at z = -0.265 from the tables:
P = 0.6045
c. Why can the normal distribution be used in part (b),
even though the sample size does not exceed 30?
I believe this is because we are given the population
standard deviation sigma rather than the sample standard deviation. So we can
use the z test.
R is between S and T, so this implies that R is on line ST and we can say
SR+RT = ST
plug in the given expressions to get
(-2x+24)+(4x+10) = 4x+12
Now solve for x
(-2x+24)+(4x+10) = 4x+12
-2x+24+4x+10 = 4x+12
2x+34 = 4x+12
2x+34-2x = 4x+12-2x
34 = 2x+12
34-12 = 2x+12-12
22 = 2x
2x = 22
2x/2 = 22/2
x = 11
If x = 11, then RS is,
RS = -2*x+24
RS = -2*11+24
RS = -22+24
RS = 2
-------------------------------------
-------------------------------------
Answers:
x = 11 and RS = 2
Answer:
B. 9(4+8)
Step-by-step explanation:
Start by simplifying the original expression:
36+72=108
Now, let's try each answer choice:
A. 18(2+3)
Add the numbers in the parentheses:
18(5)=90 --> wrong!
B. 9(4+8)
Add the numbers in the parentheses:
9(12)=108 --> correct!
C. 3(10+24)
Add the numbers in the parentheses:
3(34)=102 --> wrong!
D. 2(19+27)
Add the numbers in the parentheses:
2(46)=92 --> wrong!
Therefore the answer is B. 9(4+8)
Solving using the quadratic formula x = +-4
