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JulsSmile [24]
3 years ago
12

A survey was done that asked people to indicate whether they preferred to swim in a pool or in an ocean. The results are shown i

n the two-way table.
If 609 people were surveyed, what is the relative frequency (rounded to the nearest hundredth) of a person over 30 years old who prefers to swim in the ocean?

hint: you will have to divide the part by the whole

Question 2 options:

.54


.30


.57


.09

Mathematics
2 answers:
posledela3 years ago
8 0
Given:
                                 Pool swimming        Ocean swimming    Total
Age 30 and younger             228                         54                 282
Over 30 years old                142                        185                 327
total                                    370                        239                609

Frequencies:
                                Pool swimming        Ocean swimming    Total
Age 30 and younger     228/609 = 0.37         54/609 = 0.09       282/609 = 0.46
Over 30 years old        142/609 = 0.23       185/609 = 0.30       327/609 = 0.54
<span>total                           370/609 = 0.61       239/609 = 0.39       609/609 = 1.00</span>

<span>The relative frequency (rounded to the nearest hundredth) of a person over 30 years old who prefers to swim in the ocean is 0.30</span>    
hichkok12 [17]3 years ago
6 0

Answer: the answer is divide 185 by 609

I chose Divide 239 by 609 and it was the wrong answer.

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Graphic is showed in the figure below

Step-by-step explanation:

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Step-by-step explanation:

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Limit as x approaches 9 of x^2 -81/sqrt of x - 3

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= lim x -> 9 2x/1/2√x

Substitute x = 9

= 2(9)/1/2√9

=18/1/(2(3)

=18 × 6/1

= 108

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