The picture shows the answer I got!
Also! For the future, I use a site called math way and it is great for problems like these.
Also! For the future, I use a site called math way and it is great for problems like these.
Beach ball with diameter of 24 inches what is the volume of air in the beach ball to the nearest of a cubic foot?
1 answer:
Answer:
4 ft³
Step-by-step explanation:
From the question;
- Diameter of the beach ball is 24 Inches
We are required to determine the volume of the air in the beach ball;
- We need to know that the volume of the air is the same as the volume of the beach ball.
- The ball is in the shape of a sphere
Radius = 12 inches
But, 1 ft = 12 inches
Thus, radius = 1 ft
Therefore;
Taking pi as 3.142
Volume = 4/3 × 3.142 × 1³
= 4.189 ft³
= 4 ft³ (nearest cubic foot)
Therefore, the volume of the air in the beach ball is 4 ft³
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(2x -3)^4
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Answer:
20 percent
Step-by-step explanation:
Percentage score = × 100
= × 100
= <u>20 percent</u>
Answer:
(a) 0.94
(b) 0.20
(c) 90.53%
Step-by-step explanation:
From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let be the probability that a bearing meets the specification.
So,
Sample size, , is large.
Let X represent the number of acceptable bearing.
Convert this to a normal distribution,
Mean:
Variance:
(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.
So,
Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.
.
So, the probability that a given shipment is acceptable is
Hence, the probability that a given shipment is acceptable is 0.94.
(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.
Denote the probability od acceptance of a shipment by .
The total number of shipment, i.e sample size,
Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, , and variance,
.
Mean:
Variance:
.
In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.
.
So, the probability that a given shipment is acceptable is
Hence, the probability that a given shipment is acceptable is 0.20.
(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).
The area right to the z-score=0.99
and the area left to the z-score is 1-0.99=0.001.
For this value, the value of z-score is -3.09 (from the z-score table)
Let, be the required probability of acceptance of one shipment.
So,
On solving
Again, the probability of acceptance of one shipment, , depends on the probability of meeting the thickness specification of one bearing.
For this case,
The area right to the z-score=0.97790
and the area left to the z-score is 1-0.97790=0.0221.
The value of z-score is -2.01 (from the z-score table)
Let be the probability that one bearing meets the specification. So
On solving
Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.