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OleMash [197]
3 years ago
8

The five-number summary for a data set is given by min = 5, Q1 = 18, median = 20, Q3 = 40, max = 75. If you wanted to construct

a boxplot for the data set (that is, one that would show outliers, if any existed), what would be the maximum possible length of the right-side "whisker"? * 2 points 53 35 55 33 45
Mathematics
1 answer:
NeTakaya3 years ago
7 0

Answer:

33

Step-by-step explanation:

Values beyond the whiskers of the boxplot are called the outliers :

For a right side Outlier :

Values > Q3 ± 1.5(IQR) ;

IQR = Q3 - Q1)

From the data given:

IQR = (40 - 18) = 22

Hence ; the maximum value of right side whisker will be :

1.5(22) = 33

You might be interested in
Rewrite the form In exponential form:<br> Log100 = x
andrey2020 [161]

Answer:

10^x=100

Step-by-step explanation:

You know how subtraction is the <em>opposite of addition </em>and division is the <em>opposite of multiplication</em>? A logarithm is the <em>opposite of an exponent</em>. You know how you can rewrite the equation 3 + 2 = 5 as 5 - 3 = 2, or the equation 3 × 2 = 6 as 6 ÷ 3 = 2? This is really useful when one of those numbers on the left is unknown. 3 + _ = 8 can be rewritten as 8 - 3 = _, 4 × _ = 12 can be rewritten as 12 ÷ 4 = _. We get all our knowns on one side and our unknown by itself on the other, and the rest is computation.

We know that 3^2=9; as a logarithm, the <em>exponent</em> gets moved to its own side of the equation, and we write the equation like this: \log_3{9}=2, which you read as "the logarithm base 3 of 9 is 2." You could also read it as "the power you need to raise 3 to to get 9 is 2."

One historical quirk: because we use the decimal system, it's assumed that an expression like \log1000 uses <em>base 10</em>, and you'd interpret it as "What power do I raise 10 to to get 1000?"

The expression \log100=x means "the power you need to raise 10 to to get 100 is x," or, rearranging: "10 to the x is equal to 100," which in symbols is 10^x=100.

(If we wanted to, we could also solve this: 10^2=100, so \log100=2)

6 0
3 years ago
what is the image of the point ( -4, 8) after a rotation of 90° counterclockwise about the origin ​
lesya [120]

Answer:(-8,-4)

Step-by-step explanation:

5 0
3 years ago
$14.65 plus what equals $65.00?
ziro4ka [17]

79.65

I used a calculator

5 0
2 years ago
Read 2 more answers
#54 Simplify and write the answers using positive exponents only.
kramer
Gg easy

remember
(x^m)/(x^n)=x^(m-n)
and
(x^m)^n=x^(mn)
and
(xyz)^m=(x^m)(y^m)(z^m)
and
(m/n)^a=(m^a)/(n^a)
and
x^-n=1/(x^n)

so
( \frac{6mn^{-2}}{3m^{-1}n^2} )^{-3}=
( \frac{6}{3} )^{-3}( \frac{m}{m^{-1}} )^{-3}( \frac{n^{-2}}{n^2} )^{-3}=
(2^{-3})((m^2)^{-3})((n^{-4})^{-3})=
( \frac{1}{2^3})(m^{-6})(n^{12})=
( \frac{1}{8} )( \frac{1}{m^6})(n^{12})=
\frac{n^{12}}{8m^6}

8 0
3 years ago
find the work done by a person pushing a 40 pound box up an incline that is 30 degrees above the horizontal and 100 feet long (i
tankabanditka [31]

The work done in pushing the box is 64000 lb-ft²/s²

<h3>How to calculate the work done by the person?</h3>

Since person pushing a 40 pound box up an incline that is 30 degrees above the horizontal and 100 feet long, the work done, W is

W = mgh where

  • m = mass of box = 40 lb,
  • g = acceleration due to gravity = 32 ft/s² and
  • h = height of incline = LsinФ where
  • L = length of incline = 100 ft and
  • Ф = angle of incline = 30°

So, W = mgh

W = mgLsinФ

So, substituting the values of the variables into the equation, we have

W = mgLsinФ

W = 40 lb × 32 ft/s² × 100 ftsin30°

W = 1280 lb-ft/s² × 100 ft × 0.5

W = 1280 lb-ft/s² × 50 ft

W = 64000 lb-ft²/s²

So, the work done in pushing the box is 64000 lb-ft²/s²

Learn more about work done here:

brainly.com/question/25970931

#SPJ1

3 0
2 years ago
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