Answer:
Mean :
a) Find the probability that an individual distance is greater than 204.30 cm.
We are supposed to find P(Z>204.30)
x = 204.30
Formula :
Substitute the value sin the formula :
P(Z>204.30)=1-P(Z<204.30)
Refer the z table
P(Z>1.13253)=1-P(Z<1.13253)
P(Z>1.13253)=1-0.8708
P(Z>1.13253)=0.1292
The probability that an individual distance is greater than 204.30 cm is 0.1292.
b) Find the probability that the mean for 20 randomly selected distances is greater than 192.80 cm
We are supposed to find P(Z>192.80)
x = 192.80
Formula :
Substitute the value sin the formula :
P(Z>192.80)=1-P(Z<192.80)
Refer the z table
P(Z>-0.265)=1-P(Z<-0.265)
P(Z>-0.265)=1-0.3974
P(Z>-0.265)=0.6026
The probability that the mean for 20 randomly selected distances is greater than 192.80 cm is 0.6026
c) Why can the normal distribution be used in part (b), even though the sample size does not exceed 30
The normal distribution is used because the original population has normal distribution.
Answer
They will be able to make 18 goody bags with 24 extra toys.
Step-by-step explanation:
The least number is 18. So they will only be able to put 18 of each other toy to fill the bags with all 3 toys. So the extra toys with be the remainders.
Answer: -6
-6•-7=42
Negative times a negative equals a positive
5x + 25y = 60....multiply by -3
3x + 15y = 42....multiply by 5
-------------------
-15x - 75y = - 180 (result of multiplying by -3)
15x + 75y = 210 (result of multiplying by 5)
-------------------add
0 = 30...incorrect.....this system has no solutions
Answer:
Angle 8
Step-by-step explanation:
Refer to the attached picture