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The data of players of volleyball tournament, the mean is 56.5, median is 53, range is 56 and interquartile range is 20

The mean of the data is the average value of the given data. The mean of the data is the ratio of the sum of all the values of data to the total number of values of data.

The median of the data is the middle value of the data set when it arrange in ascending or descending order.

Mode of a data set is the value, which occurs most times for that data set. The value which has the highest frequency in the given set of data is known as the mode of that data set.

The stem-and-leaf plot shows the number of digs for the top 15 players at a volleyball tournament. The table is attached below.

The data according to the table is,

41,41,43,43,45,50,52,53,54,62,63,63,67,75,97

[tex]\mu=\dfrac{41+41+43+43+45+50+52+53+54+62+63+63+67+75+97}{15}\\\mu=\dfrac{849}{15}\\\mu=56.5[/tex]

The arranged data according to the table is,

41,41,43,43,45,50,52,53,54,62,63,63,67,75,97

Here, the middle value is 53. Thus, the median is 53.

Mode is the data which occur most offen. Here the order of modes from least to greatest is,

41   41   43   43   45   50   52   53   54   62   63   63   67   75   97  

Lowest value in data is 41 and highest is 97. Thus, the range is,

[tex]r=97-41\\r=56[/tex]

The third quartile of data is 63 while first quartile is 43. Thus, the interquartile range is.

[tex]IQR=63-23\\IQR=20[/tex]

Thus, the data of players of volleyball tournament, the mean is 56.5, median is 53, range is 56 and interquartile range is 20.

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Answer:

B, C, E

Step-by-step explanation:

Exponent Rules

[tex]a^b \cdot a^c=a^{b+c}[/tex]

[tex]\dfrac{a^b}{a^c}=a^{b-c}[/tex]

[tex]a^n \cdot b^n=(ab)^n[/tex]

[tex]a^0=1[/tex]

[tex]\dfrac{1}{a^b}=a^{-b}[/tex]

[tex](a^b)^c=a^{bc}[/tex]

[tex]\textsf{A}\quad42^{-3} \cdot 42^8=42^{-3+8}=42^5[/tex]

[tex]\textsf{B}\quad \dfrac{42^{-7}}{42^{-2}}=42^{-7-(-2)}=42^{-5}[/tex]

[tex]\textsf{C}\quad \dfrac{6^{-5}}{7^5}=6^{-5} \cdot 7^{-5}=(6 \cdot 7)^{-5}=42^{-5}[/tex]

[tex]\textsf{D}\quad 42^0 \cdot 42^5=1 \cdot 42^5=42^5[/tex]

[tex]\textsf{E}\quad \left(\dfrac{1}{42^{-5}}\right)^{-1}=(42^5)^{-1}=42^{-5}[/tex]

[tex]\textsf{F}\quad \left(\dfrac{6^{-5}}{7^5}\right)^{-1}=\left(6^{-5} \cdot 7^{-5}\right)^{-1}=\left(\left(6 \cdot 7\right)^{-5}\right)^{-1}=42^{-5(-1)}=42^5[/tex]

Check one by one

#A

[tex]\\ \rm\Rrightarrow 42^{-3}42^8=42^{-3+8}=42^5[/tex]

#B

[tex]\\ \rm\Rrightarrow \dfrac{42^{-7}}{42^{-2}}=42^{-7+2}=42^{-5}\checkmark[/tex]

#C

[tex]\\ \rm\Rrightarrow \dfrac{6^{-5}}{7^5}=6^{-5}7^{-5}=42^{-5}\checkmark[/tex]

#d

[tex]\\ \rm\Rrightarrow 42^0.42^5=42^{0+5}=42^5[/tex]

#e

[tex]\\ \rm\Rrightarrow \left(\dfrac{1}{42^{-5}}\right)^{-1}=42^{-5}\checkmark[/tex]

#f

[tex]\\ \rm\Rrightarrow \left(\dfrac{6^{-5}}{7^5}\right)^{-1}[/tex]

[tex]\\ \rm\Rrightarrow (42^{-5})^{-1}[/tex]

[tex]\\ \rm\Rrightarrow 42^5[/tex]

Answer:

[tex]log_{12}(\frac{1}{4})[/tex]

Step-by-step explanation:

We start with our equation that the change of base formula was applied to:

[tex]\frac{log\frac{1}{4} }{log12}[/tex]

The change of base formula describes taking the base (b) of the logarithm and setting it as the denominator while taking the remaining number (a) and setting that as the numerator.

If we reverse the formula, we end up with the supplied answer.  

[tex]log_{12}(\frac{1}{4})[/tex]

Answer:

Other guy is correct

Step-by-step explanation:

The number of ants that Chet's ant farm have based on the median given is 96 ants.

Your information is incomplete. Therefore, the numbers will be assumed. Let's assume that the numbers of ants in each column is 12.

In this case, when Chet's ant farm in the data, the median is 8. Therefore, the number of ants will be:

= 12 × 8

= 96 ants.

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Answer:

  see attached

Step-by-step explanation:

You are asked to use the compound interest formula for different values of n, the number of times per year the interest is compounded.

__

The basic formula is ...

  A = P(1 +r/n)^(nt)

for the amount in an account in which principal P earns interest at annual rate r compounded n times per year for t years.

The problem statement gives values for P, r, t, so we only need to compute the result for different values of n. With the given values filled in, the formula is ...

  A = 1000(1 +0.045/n)^(10n)

Values of n to be used are ...

The formula you will evaluate uses these values where 'n' is in the formula. Corresponding account balances are ...

__

For repetitive evaluation of the same formula, we like a calculator or spreadsheet. Many calculators have the "time value of money" formulas pre-programmed. This is shown in the second attachment, where the value that varies is "C/Y", the number of compoundings per year.

Answer: 5/26

Step-by-step explanation:Hope this is helpful

The amount of fabric needed = 2(surface area of the cone-shaped wooden frame) = 88 ft².

Surface area of a cone = πr² + πrl

Given the following dimensions of the cone-shaped wooden frame:

Amount of fabric needed = 2(surface area of the cone-shaped wooden frame) = 2(πr² + πrl)

= 2(3.14 × 2² + 3.14 × 2 × 5)

= 87.92

Amount of fabric needed ≈ 88 ft²

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Answer:

88

Step-by-step explanation:

Answer:

1 :

20×8=$160

160+55=$215

2:

25×8=$200

the first teenager earns more money after every shift

hope this helps you :D

Answer:

the train won't always go the same distance each minute and for the second one its 4 minutes

Answer:

A

C

B

Step-by-step explanation:

A) All the corners of the square touch the edge of the circle
B) Only 1 side of the square touches the circle so it forms a chord.

C) The circle is inside the square so the sides of the square form tangents (Perpendicular lines to the radius).

Answer: 8

Step-by-step explanation:

     Given:

(5x^2-x-1)-(-3x^2-2x-5)

     Rewrite:

(5x² -x -1) - (-3x² -2x -5)

     Distribute the negative:

5x² -x -1 + 3x² +2x +5

     Reorder terms:

5x² + 3x² -x +2x -1 +5

     Combine like terms:

8x² + x + 4

   The coefficient of the x² term is 8,

8x² + x + 4

The answer to #11 is: x = 15

The answer to #12 is: x = 5

Answer:

Is it not just EF?

Step-by-step explanation:

I mean BC is the bottom of the triangle so the corresponding part is just the other triangles bottom which is EF if it was just angle B it'd be just angle E but it's not it's both B and C so it's E and F

Answer: C. 51%

Step-by-step explanation:

Answer:

On the image

Step-by-step explanation:

...............

Answer:

457 Remainder 4

Step-by-step explanation:

Answer:

384.8 cubic inches.

Step-by-step explanation:

Answer: possible dividend is 736, 737, 738, 739

Step-by-step explanation: The remainder cannot be divided by 5 so it can be 1, 2, 3 and 4 added to (147 ×5)

Answer:

7.265

Step-by-step explanation:

Answer:

7.364

Step-by-step explanation:

Thats my answer.Brainliest

Answer: Given that MP \\NS , RS\\PQ and MR =NQ

MRN is a straight line with sum of angles equal to 180⁰ MQN is also a straight line triangle equal to 180⁰

Angle at R + 3 = Angle at Q +2 = 180⁰

Angle R is alternate to Q and angle 2 is alternate to 3 and both sets are equal

Therefore angle 1 is equal to 4 and this makes their third angles in the triangle equal.

Triangle MQP then becomes a perfect replica of NRS

Step-by-step explanation:

Answer:

Just for fun, the probability of rolling a die 60 times and getting no ones is

(600)(5/6)60=0.0018% .

Step-by-step explanation:

This 10 ones is the value with the highest probability of occurring. It has a probability of  (6010)(1/6)10∗(5/6)50≈13.7% .

First, expand the terms inside the bracket you will get

[tex](( \tan {}^{2} (x) + 2 \tan(x) \sin(x) + \sin {}^{2} (x) - ( \tan {}^{2} (x) - 2 \tan(x) + \sin {}^{2} (x) ) {}^{2} = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )[/tex]

[tex]( 4 \tan(x) \sin(x) ) {}^{2} = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )[/tex]

[tex]16 \tan {}^{2} (x) \sin {}^{2} (x) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )[/tex]

[tex]16 \tan {}^{2} (x) (1 - \cos {}^{2} (x) ) = 16 (\tan(x) + \sin(x) )( \tan(x) - \sin(x) )[/tex]

[tex]16( \tan {}^{2} (x) - \frac{ \sin {}^{2} (x) \cos {}^{2} ( {x}^{} ) }{ \cos {}^{2} (x) } [/tex]

[tex]16( \tan {}^{2} (x) - \sin {}^{2} (x) ) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )[/tex]

[tex]16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )[/tex]

D 2c + 4c + c

Collect Coefficients of like terms

[tex](2+4+1)\times c[/tex]

__________________________________________________

Calculate the sum or difference

[tex]7c[/tex]

__________________________________________________

Determine true/false

[tex]Determine\ whether\ 7c\ and\ 2c+4c+c\ are equivalent:[/tex]

[tex]True[/tex]

I hope this helps you

:)

Answer:

part: 17 players = 1/4

whole: 68 players in all

percent: 17 players = 25%

Step-by-step explanation:

We'll use p as a variable for the total players, alright?

So, 25% of p usually gets called back.

This season, 17 is 25% of p OR you might say 17 = 1/4 of p

in other words, p/4 = 17

multiply each side by 4 (to get the variable alone)

p = 17 * 4

p = 68

______________

part/whole = percent/100

SUBSTITUTE!

17/68 = 25%

check!

(simplify) 1/4 = 25% true!

Answer:

2.75

Step-by-step explanation:

0.3,1.5,1.5,1.5,2.6,2.9,3.8,3.8,4.2,9.1

(smaller to greater)

middle of this is 2.6 and 2.9

so, 2.6+2.9=5.5

now we divide it by 2

so, 5.5/2=2.75

Answer:

r=10%

I=1149.60

P=2874

t=(I×100)/(p×r)

(1149.60×100)/(2874×10)

114960÷28740

4 years

[tex]r = 10\%[/tex]

[tex]p = 2874[/tex]

[tex]l = 1149.60[/tex]

the numbers of year's needed for the investment.

[tex]t = \frac{l \times 100}{p \times r} [/tex]

[tex]t = \frac{1149.60 \times 100}{2874 \times 10} [/tex]

[tex]t = \frac{114960}{28740} [/tex]

[tex]t = 4[/tex]

therefore, 4 years is needed for the investment.

Answer:

3x10 to the power of -2

Step-by-step explanation:

0.03=3x0.01

0.01=10 to the power of -2