Last football season the starting quarterback of a professional football team completed just 55% of his attempted passes. The owner of this team has given the quarterback the ultimatum that he must improve his completion percentage during the off season or he will be fired. The owner plans to use a random sample of 50 passing attempts at the end of spring training to test the hypotheses where p represents the true proportion of passes this quarterback can complete after training during the off season. If the owner wants to minimize the chance of firing the quarterback when he really has improved, should he use a significance level of 0.01 or 0.10

let's solve your equation step-by-step.

24−3x=−27

step 1: simplify both sides of the equation.

24−3x=−27

24+−3x=−27

−3x+24=−27

step 2: subtract 24 from both sides.

−3x+24−24=−27−24

−3x=−51

step 3: divide both sides by -3.

−3x

−3

=

−51

−3

x=17

answer:

x=17

step-by-step explanation:

mark brainliest and have a great day!

step-by-step explanation:

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;