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3 years ago

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Mathematics

2 answers:

cluponka [151]3 years ago

6 0

Soo what do you need help with? It seems pretty easy. I can help you if you want.

meriva3 years ago

3 0

Answer:

a) Dot plot on hours of TV: Please see the picture below, the data are approximately symmetric.

b) Mean = 2 hours of TV

c) Median = 0.305 coins

Mean = 0.585 coins

d) Median

Step-by-step explanation:

a) To draw the dot plot, you need to sort the data smallest to largest, so you have:

0 1 1 2 2 2 3 3 4 4

Then you put in the draw one point for value you find. You can see in the draw one point for zero because is only once, two points for the one because there are two data with the value one, three dots for the two becase there are three datas with the value two, and two dots for numbers three and four because they are repeated twice in the dataset.

You can see that the distribution is approximately symmetric because there are approximately the same quantity of dots in each number.

b) As you have an approximately symmetric distribution on the data, you should use the mean to describe a typical number of hours of TV watched by the students.

So to calculate the mean you need to add up all the data and divide by how many data there are:

$Mean = \frac{0+1+1+2+2+2+3+3+4+4}{10}$

$Mean=\frac{22}{10}$

$Mean=2.2$

And round the result to an integer we have that Mean=2 hours of TV

c) Calculate of the median:

To calculate the median of the data that represent the total value of coins which has each student, you need to sort the data smallest to largest:

0.00 0.00 0.10 0.15 0.25 0.36 0.54 0.89 1.37 2.19

Now you need to find the data in the middle, in this case we have an odd set of data, so there are two data in the middle: 0.25 and 0.36, so to calculate the median, we need to find de mean between those numbers:

$Median=\frac{0.25+0.36}{2}$

Median = 0.305 coins

Calculate of the mean:

To calculate the mean of the data you need to add up all the numbers and divided by the number of data, that is:

$Mean=\frac{0.00+0.00+0.10+0.15+0.25+0.36+0.54+0.89+1.37+2.19}{10}$

$Mean=\frac{5.85}{10}$

Mean = 0.585 coins

d) The values of the mean and the median are so different because the majority of numbers are smaller than one, but there are two values that are greater than one, so that indicates that the dataset is not symmetric.

For that reason, we need to use the Median to find the value of a typical value of coins for these ten students.

You might be interested in

. The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does

Elenna [48]

Answer:

Dimension a = 18 , b= 36 will give a box with a square end the largest volume

Step-by-step explanation:

Given -

sum of box length and girth (distance around) does not exceed 108 inches.

Let b be the lenth of box and a be the side of square

b + 4a = 108

b = 108 - 4a

Volume of box = $area \times lenth$

= $a^2\times b$

V = $a^2\times b$

puting the value of b

V = $a^2 ( 108 - 4a )$

$V = 108a^2 - 4a^3$

To find the maximum value of V

(1) we differentiate it

$\frac{\mathrm{d} V}{\mathrm{d} a} = 216a - 12a^2$

(2) $\frac{\mathrm{d} V}{\mathrm{d} a} = 0$

$216a - 12a^2$ = 0

12a ( 18 - a ) =

a = 0 and a = 18

(3) putting the value of a if $\frac{\mathrm{d^2} V}{\mathrm{d} a^2}$ = negative then the value for a ,V is maximum

$\frac{\mathrm{d^2} V}{\mathrm{d} a^2}$ = 216 - 24a

put the value of a = 0 , $\frac{\mathrm{d^2} V}{\mathrm{d} a^2}$ = 216

put the value of a = 18 , $\frac{\mathrm{d^2} V}{\mathrm{d} a^2} =$ negative

for the value of a =18 V gives maximum value

Max volume = $108\times18^2 - 4\times18^3$

= 11664

a = 18 , b = 108 - 4a = $108 - 4\times 18$ = 36

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%28-60%29-3x%5E3-33x%2B24%20x%5E%7B2%7D%20" id="TexFormula1" title="(-60)-3x^3-33x+24 x^{2} "

soldi70 [24.7K]

First you want to reduce the brackets
Next you want to factor out the common number 3
Next you want to factor 20 + x^3 + 11x - 8x^2 by using the polynomial division method/technique
Lastly you just factor x^2 - 9x + 20 to get the answer.

Yes you are correct, the solutions are 1, 4, and 5.

6 0

3 years ago

What's the answer to this question

sesenic [268]

C. 110. This is because all triangles add up to equal 180*. So you add the two angles you have and subtract from 180.

5 0

3 years ago

5/6 + 2/3 + 1/4 = ???

jeka94

7/4 is the answer to your question

6 0

2 years ago

PLEASE HELP! The table shows the number of championships won by the baseball and softball leagues of three youth baseball divisi

Irina18 [472]

Answer:

Question 1: $P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16}}{ \frac{4}{16}} = \frac{1}{2}$

Question 2:

A. $P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

B. $P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

C. $P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

Step-by-step explanation:

Conditional probability is defined by

$P(A|B)= \frac{P(A and B)}{P(B)}$

with P(A and B) beeing the probability of both events occurring simultaneously.

Question 1:

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16} }{ \frac{4}{16} } = \frac{1}{2}$

Question 2.A:

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

Question 2.B:

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( Z and B)= \frac{ 1 }{ 16 }[/tex]

By definition,

$P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

Question 3.B

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

then

$P (Y or Z) = P(Y) + P(Z) = \frac{6}{16}$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

$P((YorZ) and B)= \frac{3}{16}$

By definition,

$P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

3 0

3 years ago