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3 years ago

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Karen is typing a report.she typed 3/8of the pages in the report in the morning and 2/5 of the pages in the afternoon.what fract

ion more of the pages did she type in the morning?

1 answer:

Karolina [17]3 years ago

4 0

Simple...

you have $\frac{3}{8}$ and $\frac{2}{5}$

Just subtract...but first find a common denominator...

Use 40 as the common denominator...

$\frac{3}{8} * \frac{5}{5} = \frac{15}{40}$
and
$\frac{2}{5} * \frac{8}{8} = \frac{16}{40}$

So, subtract,

$\frac{15}{40}
-
 \frac{16}{40}$

= $- \frac{1}{40}$

Thus, your answer.

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On a coordinate plane, a line is drawn from point K to point J. Point K is at (160, 120) and point J is at (negative 40, 80). Wh

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The x- and y- coordinates of point P on the directed line segment from K to J such that P is Three-fifths the length of the line segment from K to J is (85, 105).

Given

On a coordinate plane, a line is drawn from point K to point J. Point K is at (160, 120) and point J is at (negative 40, 80).

<h3>Coordinates</h3>

The coordinates point any point can be found by using the following formula.

$\rm X=\dfrac{m}{m+n}(x_2-x_1)+x_1\\\\ Y=\dfrac{m}{m+n}(y_2-y_1)+y_1$

The x- and y- coordinates of point P on the directed line segment from K to J such that P is Three-fifths the length of the line segment from K to J is;

$\rm X=\dfrac{m}{m+n}(x_2-x_1)+x_1\\\\ X = \dfrac{3}{3+5}(-40-160)+160\\\\ X = 3 \times (-25)+160\\\\\X = -75+160\\\\ X = 85\\\\\ Y=\dfrac{m}{m+n}(y_2-y_1)+y_1\\\\ Y=\dfrac{3}{3+5}(80-120)+120\\\\Y =- 3\times 5 +`120\\\\Y=-15+120\\\\Y=105$

Hence, the x- and y- coordinates of point P on the directed line segment from K to J such that P is Three-fifths the length of the line segment from K to J is (85, 105).

To know more about co-ordinates click the link given below.

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2 years ago

Solve the literal equation for the given variable.<br> y = x + 12; x

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Is this a graphing question or…

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A student wrote the following proof. Evaluate the student’s answer.

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2 years ago

8x - 5 = 28x + 10, what does x equal?

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2 years ago

A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d

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Answer:

(10) Person B

(11) Person B

(12) $P(5\ or\ 6) = 60\%$

(13) Person B

Step-by-step explanation:

Given

Person A $\to$ 5 coins (records the outcome of Heads)

Person $\to$ Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

$Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}$

$n(Head) = 32$

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

Next, we list out the highest in each toss (sorted)

$Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}$

$n(Dice) = 30$

Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Head)}$

$Pr(5) = \frac{1}{32}$

$Pr(5) = 0.03125$

From person B list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Dice)}$

$Pr(5) = \frac{8}{30}$

$Pr(5) = 0.267$

<em>From the above calculations: </em> $0.267 > 0.03125$ <em> Hence, person B is more likely to get 5</em>

Question 11: Person with Higher median

For person A

$Median = \frac{n(Head) + 1}{2}th$

$Median = \frac{32 + 1}{2}th$

$Median = \frac{33}{2}th$

$Median = 16.5th$

This means that the median is the mean of the 16th and the 17th item

So,

$Median = \frac{3+2}{2}$

$Median = \frac{5}{2}$

$Median = 2.5$

For person B

$Median = \frac{n(Dice) + 1}{2}th$

$Median = \frac{30 + 1}{2}th$

$Median = \frac{31}{2}th$

$Median = 15.5th$

This means that the median is the mean of the 15th and the 16th item. So,

$Median = \frac{5+5}{2}$

$Median = \frac{10}{2}$

$Median = 5$

<em>Person B has a greater median of 5</em>

Question 12: Probability that B gets 5 or 6

This is calculated as:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

From the sample space of person B, we have:

$n(5\ or\ 6) =n(5) + n(6)$

$n(5\ or\ 6) =8+10$

$n(5\ or\ 6) = 18$

So, we have:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

$P(5\ or\ 6) = \frac{18}{30}$

$P(5\ or\ 6) = 0.60$

$P(5\ or\ 6) = 60\%$

Question 13: Person with higher probability of 3 or more

Person A

$n(3\ or\ more) = 16$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}$

$P(3\ or\ more) = \frac{16}{32}$

$P(3\ or\ more) = 0.50$

$P(3\ or\ more) = 50\%$

Person B

$n(3\ or\ more) = 28$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}$

$P(3\ or\ more) = \frac{28}{30}$

$P(3\ or\ more) = 0.933$

$P(3\ or\ more) = 93.3\%$

By comparison:

$93.3\% > 50\%$

Hence, person B has a higher probability of 3 or more

7 0

2 years ago