Devon wants to build a ramp with the dimensions shown. How much wood does he need?
1 answer:
The image of the ramp with dimensions is missing, so i have attached it.
Answer:
680 in² of wood is needed.
Step-by-step explanation:
The way to find how much wood would be needed by devon would be to find the total surface area of the ramp.
From the attached image,
Let's find the area of the 2 triangles first;
A1 = 2(½bh) = bh = 15 x 8 = 120 in²
Area of the slant rectangular portion;
A2 = 17 x 14 = 238 in²
Area of the base;
A3 = 15 × 14 = 210 in²
Area of vertical rectangle;
A4 = 8 × 14 = 112 in²
Total Surface Area = A1 + A2 + A3 + A4 = 120 + 238 + 210 + 112 = 680 in²
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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.