What side do DEC and EDB have in common?
2 answers:
<u><em>Answer:</em></u>
DE is the common side in the two triangles
<u><em>Explanation:</em></u>
The side of the triangle is named based on its start and end points
A triangle is named based on its three vertices where each two vertices would join forming a side
<u>For ΔDEC, the sides are:</u>
DE, EC and DC
<u>For ΔEDB, the sides are:</u>
ED, DB and EB
By comparing the two triangles, we would find that DE is the common side in both triangles
Note: The visual illustration is in the attached image
Hope this helps :)
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Answer:
The angle between [A_F] and the base of the cone = 68.2°
The area of the base of the cone ≈ 12.57 m²
Step-by-step explanation:
The given parameters are;
The height of the cone = 5 m
The base radius of the cone = 2 m
The angle which the AC = 120°
Therefore, we have;
The angle between [A_F] and the base of the cone = The angle between [CF] and the base of the cone
The angle between [CF] and the base of the cone = tan⁻¹(5/2) = tan⁻¹(2.5) ≈ 68.2°
∴ The angle between [A_F] and the base of the cone = The angle between [CF] and the base of the cone = 68.2°
The angle between [A_F] and the base of the cone = 68.2°
The area of the base of the cone = π × r² = π × 2² = 4·π ≈ 12.57
The area of the base of the cone ≈ 12.57 m².
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!