The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. an article reports the following summary data on intake for a sample of males age 65−74 years: n = 114, x = 11.9, and s = 6.42.

does this data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance? (use α = 0.05.)calculate the test statistic and determine the p-value. (round your test statistic to two decimal places and your p-value to four decimal places.)

Data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance.

Step-by-step explanation:

Let mu be the average daily zinc intake in the population of all males age 65−74

: mu= 15 mg/day

: mu< 15 mg/day

Test statistic can be calculated using the equation:

z= where

Then z= ≈ -5.15

Since p-value≈0.0001 < 0.05 is very low and smaller than significance level, we can reject the null hypothesis.

0.0076, if you meant 4.125 divided by 0.03125 it would be 132

step-by-step explanation:

first you have to move the decimal in 0.03125 over until its behind the 5 which is 5 places over. now the number is 3125

then you have to move the decimal in 4.125, 5 places over. now the number is 4,112,500.

now you divide like you would normally would and get 0.0076

the correct answer option is (2x-3)(2x+3).

step-by-step explanation:

we are given the following expression and we are to factorize it:

we are to find such factors that when multiplied they give a result of 36 (9 x 4 = 36) and when added they give a result of 0.

the two such factors of 36 are negative six and postive six.

factorizing the given expression to get:

therefore, the correct factorization of is (2x-3)(2x+3).