There is a 60% that your math teacher will give a pop quiz tonight and, due to the fact that you have a soccer game, there is an 85% chance you will not be able to do your math homework. you decide that you will do the homework if the probability of there being a pop quiz and your not doing your homework is greater than 60%. what do you do and why?

Step-by-step explanation:

Let A be the event that your math teacher will give a pop quiz tonight

and B the event that you have a soccer game

GIven that P(A) = 60% = 0.60

P(B)=85% =0.85

We have A and B are independent since pop quiz and soccer game have no association

Hence P(AB) = P(A) P(B)

THus probability for both pop quiz and soccer game = 0,60(0.85)

=0.51

Obviously 0.51 <0.60

So you need not do the homework

1/4 because there is 4 grades.

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;