1.) 3/4
2.) It snowed 1/4 more on Tuesday than it did on Monday.
Answer:
C. n=23; p^=0.5
Step-by-step explanation:
Normal distribution is symmetrical about the mean.
So, p should be close to ½
Answer:
The radius of the scoop is r = 3.1 cm
Step-by-step explanation:
Since 3 gallons yields 90 scoops, and 1 gallon = 3785 cm³.
3 gallons = 3 × 3785 cm³ = 11355 cm³
So we have 11355 cm³ in 3 gallons which is also the volume of 90 scoops.
Since 90 scoops = 11355 cm³, then
1 scoop = 11355 cm³/90 = 126.2 cm³
Now, if each scoop is a sphere, the volume is given by V = 4πr³/3 where r is the radius of the scoop. Since we need to find the radius of the scoop, r, making r subject of the formula, we have
r = ∛(3V/4π)
Substituting V = 126.2 cm³, we have
r = ∛(3× 126.2 cm³/4π)
= ∛(378.6 cm³/12.57)
= ∛30.13 cm³
= 3.1 cm
So, the radius of the scoop is r = 3.1 cm
Answer:
$189
Step-by-step explanation:
35% of 140 = 49
140 + 149 = 189
Answer:
(0.5848 ; 0.6552)
We are confident that about 58% to 66% of sea foods in the country are Mislabelled.
No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.
Step-by-step explanation:
Sample size, n = 51
p = 0.62
1 - p = 1 - 0.62 = 0.38
n = 515
Confidence level = 90% = Zcritical at 90% = 1.645
Confidence interval = (p ± margin of error)
Margin of Error = Zcritical * sqrt[(p(1-p))/n]
Margin of Error = 1.645 * sqrt[(0.62(0.38))/515]
Margin of Error = 1.645 * 0.0214
Margin of Error = 0.035203
Lower boundary = (0.62 - 0.035203) = 0.584797
Upper boundary = (0.62 + 0.035203) = 0.655203
(0.5848 ; 0.6552)
We are confident that about 58% to 66% of sea foods in the country are Mislabelled.
No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.
