X=? Please help!!! Will give brainliest
1 answer:
You might be interested in
Answer:
The two airplanes are about 330miles apart.
Step-by-step explanation:
The diagram interpreting the question has been attached to this response.
As shown in the diagram,
i. the airplanes leave at point C.
ii. at 2.00pm the first and second airplanes are at points A and B respectively, where they are 312miles and 487miles away from the starting point C in directions due north and N42E from the point C.
iii. the points A, B and C form a triangle with sides a, b and c.
To solve for the value of c which is the distance between the two planes at 2.00pm, the cosine rule is used.
c² = a² + b² - 2abcosC --------------(i)
where;
b = 312miles
a = 487miles
C = 42°
Substitute these values into equation (i) and solve as follows;
c² = (487)² + (312)² - 2(312)(487)cos(42)
c² = (237169) + (97344) - 303888cos(42)
c² = (237169) + (97344) - 303888(0.7431)
c² = 334513 - 225819.1728
c² = 108693.8272
<em>Take the square root of both sides</em>
√c² = √108693.8272
c = 329.69
c ≅ 330miles
Therefore, the two airplanes are far apart by 330miles
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!