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Dominik [7]
2 years ago
9

A survey of 2237 adults ages 18 and over asked which sport is their favorite. The results are shown in the figure. What is the p

robability that an adult chosen at random prefers auto racing? Round you answer to the nearest whole percent.

Mathematics
1 answer:
Sergeu [11.5K]2 years ago
6 0

This question is incomplete because it lacks the appropriate figure. Find attached to this answer the diagram of the figure.

Answer:

8%

Step-by-step explanation:

The formula for Probability =

Numbers of possible outcomes ÷ Number of events.

From the diagram, Number of Possible outcomes is given as = Number of Adults that prefer auto racing = 179 adults

Number of events is obtained from the questions as the Number of Adults = 2237 adults

Probability that an adult chosen at random prefers auto racing = 179 adults /2237 adults

= 0.0800178811

Rounding my answer to the nearest whole percent, the probability that an adult chosen at random prefers auto racing

= 0.0800178811 × 100

= 8%

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With reference to the figure match the angles and arcs to their measures
photoshop1234 [79]
Angle BAC = (180 - 114) degrees = 66 degrees        [angle on a straight line]

Angle OCA = Angle OBA = 90 degrees                     [angle at the point where the tangent and the radius meet]

Thus, the measure of arc BC = (360 - 66 - 90 - 90) degrees = 114 degrees      [sum of interior angles of a quadrilateral]


Angle CDE = (180 - 124) degrees = 56 degrees        [angle on a straight line]

Angle OCD = Angle OED = 90 degrees                     [angle at the point where the tangent and the radius meet]

Thus, the measure of arc CE = (360 - 56 - 90 - 90) degrees = 124 degrees     [sum of interior angles of a quadrilateral]

Given that the measure of arc BC is 114 degrees and the measure of arc CE is 124 degrees, thus the measure of arc BE = (360 - 114 - 124) degrees = 122 degrees                                                      [angle at a point]

Angle OBF = Angle OEF = 90 degrees                     [angle at the point where the tangent and the radius meet]

Thus, angle BFE = (360 - 122 - 90 - 90) degrees = 58 degrees.
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Write a number sentence that will calculate a reasonable estimate for the quotient of 3325÷5​
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3325/5=665

Step-by-step explanation:

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If four is half of five and two thirds of six is nine. what is thirty-two?
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In professional basketball games during 2009-2010, when Kobe Bryant of the Los Angeles Lakers shot a pair of free throws, 8 time
ziro4ka [17]

Answer:

Is plausible that the successive throws are independent

Step-by-step explanation:

1) Table with info given

The observed values are given by the following table

__________________________________________________

First shot          Made          Second shot missed           Total

__________________________________________________

Made                  152                       33                                185

Missed                37                         8                                  45

__________________________________________________

Total                    189                       41                                230

2) Calculations and test

We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:

Null hypothesis: Independence between two successive free throws

Alternative hypothesis: No Independence between two successive free throws

_____________________________________________________

First shot                    Made                                    Second shot missed

_____________________________________________________

Made                  189(185)/230=152.0217                41(185)/230=32.9783

Missed                189(45)/230=36.9783                  41(45)/230=8.0217

_____________________________________________________

On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.

3) Calculate the chi square statistic

The statistic for this case is given by:

\chi_{cal}^2 =\sum \frac{(O_i -E_i)}{E_i}

Where O represent the observed values and E the expected values. Replacing the values that we got we have this

\chi_{cal}^2 =\frac{(152-152.0217)^2}{152.0217}+\frac{(33-32.9783)^2}{32.9783}+\frac{(37-36.9783)^2}{36.9783}+\frac{(8-8.0217)^2}{8.0217}=0.000003098+0.00001428+0.00001273+0.0.00005870=0.00008881

Now with the calculated value we can find the degrees of freedom

df=(r-1)(c-1)=(2-1)(2-1)=1 on this case r means the number of rows and c the number of columns.

Now we can calculate the p value

p_v =P(\chi^2 >0.00008881)=0.9925

On this case the pvalue is a very large value and that indicates that we can fail to reject the null hypothesis of independence. So is plausible that the successive throws are independent.

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2 years ago
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