Answer:
a) 0.5047
b) 0.5978
c) Yes
Step-by-step explanation:
Given:
Mean, u = 0.8548
Standard deviation = 0.0512
Sample mean, X' = 0.8542
a) If 1 candy is randomly selected, the probability that it weighs more than 0.8542 would be:
From standard normal table, NORMSTD(-0.0117) = 0.4953
P(z > -0.0117) = 1 - 0.4953 = 0.5047
Probability = 0.5047
b) If 447 candies are randomly selected the probability that their mean weight is at least 0.8542:
Here, we are to find the probability that the men weight is greater or equal to 0.8542
From standard normal table, NORMSTD(-0.24776) = 0.40216
P(z > -0.0117) = 1 - 0.40216 = 0.5978
Probability = 0.5978
c) Yes, it seems the candy company is providing consumers with the amount claimed on the label, because the probability of getting a sample mean of 0.8542 or greater when 447 candies are selected is not exceptionally small
Subtract 12 seconds from 11.7 seconds
The Answer is .3
We know the following:
Cylinder volume: V₁ = π r² h
Ball (sphere) volume:V₂ =
π r³
where:
V - volume
r - radius of base of cylinder and diameter of ball
h - height of cylinder.
R = 13 cm ⇒ r = 13 ÷ 2 = 6.5
π = 3.14
a) Since balls touch all sides of cylinder (as shown in image), it can be concluded that height of cylinder is equal to sum of diameters of 3 balls and that radius of base of cylinder is equal to radius of ball:
h = 3 × r = 3 × 13 cm = 39 cm
r = 6.5 cm
So,
V₁ = <span>π r² h
</span><span>V₁ = </span>3.14 × (6.5 cm)² × 39 cm
V₁ = 5,173.9 cm³
b. The total volume of three balls is the sum of volumes of each ball:
Vₐ = 3 × <span>V₂
</span>Vₐ = 3 × <span>
π r³
</span>Vₐ = 3 ×
3.14<span> (6.5 cm)³</span>
Vₐ = 3,449.3 cm³
c. Percentage of the volume of the container occupied by three balls ould be expressed as ratio of volume of three balls and volume of cylinder:
V =
×100
V =
×100
V = 0.6666 ×100
V = 66.66%
Answer:A
Step-by-step explanation just try it it bascically 31 times it self 4 times
I think the answer would be be C.