Answer:
Step-by-step explanation:
Mahala is standing and fishing on a dock that is 2 1/4 feet above the surface of the water in the pond. Her baited hook is 1.5 f
eet below the surface of the water. The surface of the water is represented by the rational number 0. Use the number line to graph the distance the dock, point D, is above the water and the hook, point H, is below the water. Make sure you label each point appropriately.
1 answer:
Answer:
The graph is shown below.
Step-by-step explanation:
Take the distance above surface of water as positive and negative for below the surface.
Given:
Distance of dock from the surface of water,
Distance of hook from the surface of water,
The surface of water is the reference point and hence it has a value of 0.
Now, on a number line, the values to the left of 0 mark is negative and values to the right are positive.
Therefore, H will lie on the left of 0 mark and D will lie on the right as shown below.
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Answer:
Step-by-step explanation:
Hello!
The variable of interest is X: ounces per cup dispensed by the cola-dispensing machine.
The population mean is known to be μ= 8 ounces and its standard deviation σ= 1.0 ounce. Assuming the variable has a normal distribution.
A sample of 34 cups was taken:
a. You need to calculate the Z-values corresponding to the top 5% of the distribution and the lower 5% of it. This means you have to look for both Z-values that separates two tails of 5% each from the body of the distribution:
The lower value will be:
= -1.648
You reverse the standardization using the formula
X[bar]= 7.72ounces
The lower control point will be 7.72 ounces.
The upper value will be:
X[bar]= 8.28ounces
The upper control point will be 8.82 ounces.
b. Now μ= 7.6, considering the control limits of a.
P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)
P(Z≤(8.28-7.6)/(1/√34))- P(Z≤7.72-7.6)/(1/√34))
P(Z≤7.11)- P(Z≤0.70)= 1 - 0.758= 0.242
There is a 0.242 probability of the sample means being between the control limits, this means that they will be outside the limits with a probability of 1 - 0.242= 0.758, meaning that the probability of the change of population mean being detected is 0.758.
b. For this item μ= 8.7, the control limits do not change:
P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)
P(Z≤(8.28-8.7)/(1/√34))- P(Z≤7.72-8.7)/(1/√34))
P(Z≤-2.45)- P(Z≤-5.71)=0.007 - 0= 0.007
There is a 0.007 probability of not detecting the mean change, which means that you can detect it with a probability of 0.993.
I hope it helps!