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marta [7]
3 years ago
15

On the distant plant, Mathology, a sports area covers 7400 yodels2. How many square deckles would this be if 1 deckle = 75.9 yod

els?
Mathematics
1 answer:
STatiana [176]3 years ago
8 0

Answer:

It would be <u>97.50</u> square deckles.

Step-by-step explanation:

Given:

On the distant plant, Mathology, a sports area covers 7400 yodels².

1 deckle = 75.9 yodels.

Now, to get the square deckles.

As given, 1 deckle = 75.9 yodels.

So, to get the square deckles by using conversion factor:

<em>75.9 yodels = 1 deckle.</em>

7400 yodels² = 7400\div 75.9

                       = 97.50\ deckels ^2.

Therefore, it would be 97.50 square deckles.

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Suppose that the weights of passengers on a flight to Greenland on Frigid Aire Lines are normal with mean 175 pounds and standar
Natali5045456 [20]

Answer:

The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 175, \sigma = 22

What is the minimum weight for a passenger who outweighs at least 90% of the other passengers?

90th percentile

The 90th percentile is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. So

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{X - 175}{22}

X - 175 = 22*1.28

X = 203.16

The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds

6 0
3 years ago
ASAP please help me ​
Marizza181 [45]

Answer:

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Step-by-step explanation:

hope this helps somewhat

7 0
3 years ago
A square of side length s lies in a plane perpendicular to a line L. One vertex of the square lies on L. As this square moves a
user100 [1]

Answer:

Part (A) The required volume of the column is s^2h.

Part (B) The volume be s^2h=\frac{s^2h}{2}+\frac{s^2h}{2}.

Step-by-step explanation:

Consider the provided information.

It is given that the we have a square with side length "s" lies in a plane perpendicular to a line L.

Also One vertex of the square lies on L.

Part (A)

Suppose there is a square piece of a paper which is attached with a wire through one corner. As you blow it up it spins around on the wire.

This square moves a distance h along​ L, and generate a​ corkscrew-like column with square​.

The cross section will remain the same.

So the cross section area of original column and the cross section area of twisted column at each point will be the same.

The volume of the column is the area of square times the height.

This can be written as:

s^2h

Hence, the required volume of the column is s^2h.

Part (B) What will the volume be if the square turns twice instead of once?

If the square turns twice instead of once then the volume will remains the same but divide the volume into two equal part.

s^2h=\frac{s^2h}{2}+\frac{s^2h}{2}

Hence, the volume be s^2h=\frac{s^2h}{2}+\frac{s^2h}{2}.

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kolbaska11 [484]

Answer:

B.

Step-by-step explanation: Because when you count them down to 3/5 It's more in the negative.

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Answer:

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