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weqwewe [10]
1 year ago
12

Could anybody explain to me what both the question is asking in a simple way. For a dummy.

Mathematics
1 answer:
almond37 [142]1 year ago
5 0
Question A is asking: Considering the distribution shape (i.e skewed left, right, or symmetrical), what measure of center would be used? Explain why that measure of center would be used.

Question B is asking: What measure of variability (or change, i.e range, interquartile range, or standard deviation) would be used for this graph? Explain why this measure would be used.
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Complete the steps to add 3 2/5 +1 3/10 . 1. Write mixed numbers as improper fractions. 2. Write equivalent fractions with a com
ira [324]

Answer:

47/10 = 4 7/10

Step-by-step explanation:

3 2/5 + 1 3/10

1. Write mixed numbers as improper fractions:

17/5 + 13/10

2. Write equivalent fractions with a common denominator:

34/10 + 13/10

3. Add the numerators over the common denominator.

47/10

4. Simplify.

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3 years ago
Before every​ flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the ai
ollegr [7]

Answer:

The probability that the aircraft  is  overload = 0.9999

Yes , The pilot has to be take strict action .

Step-by-step explanation:

P.S - The exact question is -

Given - Before every​ flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The aircraft can carry 37 ​passengers, and a flight has fuel and baggage that allows for a total passenger load of 6,216 lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than 6216/37 = 168 lb. Assume that weight of men are normally distributed with a mean of 182.7 lb and a standard deviation of 39.6.

To find - What is the probability that the aircraft is​ overloaded ?

         Should the pilot take any action to correct for an overloaded aircraft ?

Proof -

Given that,

Mean, μ = 182.7

Standard Deviation, σ = 39.6

Now,

Let X be the Weight of the men

Now,

Probability that the aircraft is loaded be

P(X > 168 ) = P(\frac{x - \mu}{\sigma} > \frac{168 - \mu}{\sigma} )

                 = P( z > \frac{168 - 182.7}{39.6} )

                 = P( z > -0.371)

                 = 1 - P ( z ≤ -0.371 )

                 = 1 - P( z > 0.371)

                 = 1 - 0.00010363

                 = 0.9999

⇒P(X > 168) = 0.9999

As the probability of weight overload = 0.9999

So, The pilot has to be take strict action .

5 0
2 years ago
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