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sveticcg [70]
3 years ago
10

Kevin recorded the ages of the next 12 people who entered his grocery store. He asked his brother John to find the mean, median,

and the mode of the data set: [ 6,18,8,4,18,20,10,10,21,6,17,18]. John's results are shown. mean =156/12=13. median = 20+10/2=15, mode=10.
A. Determine whether John's answers are correct. If an answer is incorrect, explain his error and provide the correct response. Show your work.

B. The data set was expanded to include the ages of the next three people who entered the store: 10, 12, and 52. Compare the new measures of center to those in part (a), Show your work.

C. Determine the best measure of center for each data set. Justify your answer
Mathematics
1 answer:
Zepler [3.9K]3 years ago
7 0

Answer:

The correct median is 13.5

The correct mode is 18

The arithmetic mean is 13

<em>(The geometric mean is 11.4)</em>

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A shop sells 3 pairs of boots at size 6, 5 pairs at size 7. 8 pairs at
alina1380 [7]

modusAnswer:

median=8

modusmodusmodus=8

Step-by-step explanation:

6,6,6,7,7,7,7,7,8,8,8,8,8,8,8,8,9,9,9,9,11

median=[ttTTeTTextTTeTTexteTexTTeTTextTTeTTexteTex]X_{\frac{21+1}{2}}=X_11

4 0
2 years ago
In a clinical study, volunteers are tested for a gene that has been found to increase the risk for a disease. The probability th
MrRa [10]

Answer:

a) 0.984

b) 20 people

Step-by-step explanation:

a)

If The probability that a person carries the gene is 0.1, then in a sample of 20 people, 2 should carry the gene.

Now, we want to know how many samples there are with this property.

Since we have 20 elements where 18 are alike (do not carry the gene) and 2 are alike (carry the gene), we have to compute the number of permutations of 20 elements in which 18 are alike and 2 are alike. This number is

\frac{20!}{18!2!}=190

In this 190 20-tuples there are only 3 where the 2 carriers of the gene are in the first 3 places, namely

(1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

where 1=carries the gene, 0=does not carry the gene.

So there are 190-3 = 187 elements in which the first 3 elements have no 2 carriers, hence the probability that 4 or more people will have to be tested until 2 of them with the gene are detected is 187/190 = 0.984 (98.4%) rounded to three decimal places.

b)

Given that the probability that a person carries the gene is 0.1, then in a sample of 20 people, 20*0.1 = 2 should carry the gene.

4 0
3 years ago
I GIVE BRAINLIEST PLS HELP
ioda
The answer is 8
Here's why:
{ ( \frac{( {6}^{7}) \times ( {3}^{3})  }{(  {6}^{6}) \times ( {3}^{4}  ) } )}^{3}  =   \\ ( \frac{6}{3} ) ^{3}  =  \\ \frac{216}{27}  = 8
The exponents are subtracted one from another when divided.
\frac{ a ^{b} }{ {a}^{c} } =  {a}^{b - c}
We can look at the problem this way:
( \frac{6^{7} }{6 ^{6} }  \times  \frac{3^{3} }{ {3}^{4} } ) = (6^{7 - 6}  \times  {3}^{3 - 4} ) =  \\ ({6}^{1}  \times  {3}^{ - 1} )
Since we have the power of -1 on the 3, we apply this rule:
{a}^{ - b}  =  \frac{1}{ {a}^{b} }
Also this rule because we have the power of 1 on the 6:
{a}^{1}  = a
Then we get this:
(6 \times  \frac{1}{3} )^{3}  = ( \frac{6}{3} )^{3}
We apply the rule:
( \frac{a}{b}) ^{c}   =  \frac{ {a}^{c} }{ {b}^{c} }
We get this:
\frac{{6}^{3} }{ {3}^{3} } =  \frac{216}{27}  = 8
4 0
4 years ago
Divide (10m^8+4m^6+10m^5) / 2m^2
anastassius [24]

Answer:

last option at the bottom

Step-by-step explanation:

(see attached)

3 0
3 years ago
Read 2 more answers
The price of a sandwich decreases from $6 to $4. What is the percent decrease in the price of the sandwich?
Anettt [7]

Answer:

33.33%

Step-by-step explanation:

Decrease in price= Original price -- New price

= $6 - $4

= $2

Percentage decrease= $2÷$6 × 100

= 33.33%

7 0
3 years ago
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