The design team in charge of the warehouse has chosen a tile that has an area of 2.25 square feet. Which of the following situat
2 answers:
Answer:
The answer is D. none of the above
Step-by-step explanation:
To calculate the number of tiles needed you have to use following model
In the formula:
- <em>t </em>is the number of tiles needed.
- <em>x </em>is the area of the warehouse that we need to cover with tiles.
- 2,25 square feet is the area that covers one tile. (this is given data in the question).
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Answer:
$1.13
Step-by-step explanation:
Since 6 pound bag of oranges costs $6.78...
Now, let's divide 6.78 in to 6 to find our unit price.
Now we know 1 pound, (our unit price), costs...
Answer:
56.7 mi/h
Step-by-step explanation:
The formula for average speed is:
Average Speed =
Thus we have:
Average Speed = mph
<em>Rounding to the nearest tenth, we have </em>
Average Speed = 56.7 mi/h
Answer:
(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.
(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.
(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.
(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size
(e) The population mean may be larger than 75 minutes between irruption.
Step-by-step explanation:
We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.
(a) Let X = <u><em>the interval of time between the eruption</em></u>
So, X ~ Normal()
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean time between irruption = 75 minutes
= standard deviation = 20 minutes
Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)
P(X > 84 min) = P( >
) = P(Z > 0.45) = 1 - P(Z
0.45)
= 1 - 0.6736 = <u>0.3264</u>
The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.
(b) Let = <u><em>sample time intervals between the eruption</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = population mean time between irruption = 75 minutes
= standard deviation = 20 minutes
n = sample of time intervals = 13
Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( > 84 min)
P( > 84 min) = P(
>
) = P(Z > 1.62) = 1 - P(Z
1.62)
= 1 - 0.9474 = <u>0.0526</u>
The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.
(c) Let = <u><em>sample time intervals between the eruption</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = population mean time between irruption = 75 minutes
= standard deviation = 20 minutes
n = sample of time intervals = 20
Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( > 84 min)
P( > 84 min) = P(
>
) = P(Z > 2.01) = 1 - P(Z
2.01)
= 1 - 0.9778 = <u>0.0222</u>
The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.
(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.
(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.