Answer:
the probability that a sample of 4 bags will have a mean weight less than 9.8 pounds is 0.05
Step-by-step explanation:
Given the data in the question;
μ_x = 10 pound bags
standard deviation s_x = 0.24 pounds
sample size n = 4
The bag weights are normally distributed so;
p( x' less than 9.8 ) will be;
p( (x'-μ_x' / s_x') < (9.8-μ_x' / s_x') )
we know that;
μ_x' = μ_x = 10
and s_x' = s_x/√n = 0.24/√4
so; we substitute
p( z < ( (9.8 - 10) / (0.24/√4) )
p( z < -0.2 / 0.12 )
p( z < -1.67 )
{ From z-table }
⇒ p( z < -1.67 ) = 0.0475 ≈ 0.05
Therefore, the probability that a sample of 4 bags will have a mean weight less than 9.8 pounds is 0.05
Take the original price and times it by .8 (you could times it by .2 and then subtract from the original price but multiplying it by .8 is much easier) 64.9x.8= 51.92 take your discounted price and multiply it by 1.085 to account for the sales tax (you multiply by 1.085 to account for the price of the item and the tax you could times it by .085 and then add it on to the price of the item but this way it shortens the amount of steps needed) 51.92x1.085= 56.3332
Round to the nearest cent
Answer=$56.33
We are given with a bag containing 13 dark chocolates, 16 white chocolates, and 11 milk chocolates. hence the sample space is 13 + 16 + 11 equal to 40 chocolates. The <span> probability that she randomly picks a white chocolate is 16/40 or 2/5 and that she picks a milk chocolate is 11/40. Hence the probability of picking either is (16+11) /40 equal to 27/40</span>
3 blocks east and 2 blocks north should be the correct answer :)
