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nlexa [21]
3 years ago
9

Dr. Sandoval, a farm veterinarian, just weighed a particular pig and found out it is 96 kilograms. The last time she checked, th

e pig weighed 100 kilograms. What percent lighter is the pig now?
Mathematics
1 answer:
slega [8]3 years ago
4 0
4% because 100=100% -4 =96
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A high school has 36 players on the football team. the summary of the players' weights is given in the box plot. what is the int
Komok [63]

The interquartile range of the players' weights = 48 pounds.

The boxplot attached to this question is missing. It was obtained online and is attached to this solution of the question.

It should be noted that the following is true for a boxplot.

A box plot gives a visual representation of the distribution of the data, showing where most values lie and those values that greatly differ from the rest, called outliers.

The elements of the box plot are described thus;

The bottom side of the box represents the first quartile, and the top side, the third quartile. Therefore, the width of the central box represents the inter-quartile range.

The horizontal line inside the box is the median.

The lines extending from the box reach out to the minimum and the maximum values in the data set, as long as these values are not outliers. The ends of the whiskers are marked by two shorter horizontal lines.

Variables in the dataset, higher than Q3+(1.5×IQR) or lower than Q1-(1.5×IQR) are considered outliers and are usually shown using dots above the top whisker or below the bottom whisker.

So, it is evident that for this question,

First quartile = 174 pounds

Third quartile = 222 pounds

Inter Quartile Range = (Third quartile) - (First quartile)

= 222 - 174

= 48 pounds.

To know more about "interquartile"

Refer this link:

brainly.com/question/16988652

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6 0
2 years ago
Divide (13 + 6х2 +11х + 6) by (х + 2)
adoni [48]

Answer:

(13+6x²+11x+6)÷(x+2)

(6x²+11x+19)÷(x+2)

(6x²+11x+19)÷(x+2)

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Step-by-step explanation:

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4 years ago
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A poll asked 1057 American adults if they believe there was a conspiracy in the assassination of President Kennedy, and found th
AVprozaik [17]

Answer:

The 95% confidence interval of the proportion of Americans who believe in the conspiracy is  0.551<  p <  0.610

Step-by-step explanation:

From the question we are told that

   The sample size is  n =1057

   The number that believe there was a conspiracy is  k = 614

Generally the sample proportion is mathematically represented as

      \^ p =  \frac{614}{1057 }

=>   \^ p = 0.5809

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

=>   E = 1.96 * \sqrt{\frac{0.5809 (1- 0.5809)}{1057} }    

=>   E = 0.02975  

Generally 95% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

=>    0.5809 - 0.02975<  p <  0.5809 + 0.02975

=>    0.551<  p <  0.610

4 0
3 years ago
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