Astudy was conducted to determine the mean birth weight of a certain breed of kittens. consider the birth weights of kittens to be normally distributed. a sample of 45 kittens was randomly selected from all kittens of this breed at a large veterinary hospital. the birth weight of each kitten in the sample was recorded. the sample mean was 3.56 ounces, and the sample standard deviation was 0.2 ounces. what is the margin of error for a 90% confidence interval on the mean birth weight of all kittens of this breed.
b. -4/3, -1/2
we are given the following equation and we are to solve it by factorizing it:
we are to find factors of 24 such that when multiplied they give a product of 24 and when added they give a result of 11.
therefore, the correct answer option is b. -4/3, -1/2.
i will need less than 17 dollars purchase dinner.
i could spend zero to 16.99 on dinner and still not violate the inequality.
28 shorts, and 279 t-shirts.
dots is selling t-shirts for $2 a piece and has sold 279 of them that is $558.
279 shirts x 2 (dollars)= $558.
we need to see how many shorts were sold, so we are going to take the money we know didn't come from shorts and subtract it from their total sales.
$112 is left, divide 112 by 4 to see how many shorts were sold.
112/4= 28 shorts.
28 shorts and 279 t-shirts.
hope this !