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3 years ago

Find the rationalizing factor of:

Mathematics

1 answer:

nataly862011 [7]3 years ago

8 0

Answer:

Check out the answer. Hope it helps!!

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Suppose we have 3 cards identical in form except that both sides of the first card are colored red, both sides of the second car

Nikolay [14]

Answer:

probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3

Step-by-step explanation:

First of all;

Let B1 be the event that the card with two red sides is selected

Let B2 be the event that the

card with two black sides is selected

Let B3 be the event that the card with one red side and one black side is

selected

Let A be the event that the upper side of the selected card (when put down on the ground)

is red.

Now, from the question;

P(B3) = ⅓

P(A|B3) = ½

P(B1) = ⅓

P(A|B1) = 1

P(B2) = ⅓

P(A|B2)) = 0

(P(B3) = ⅓

P(A|B3) = ½

Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;

Thus;

P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]

P(B3|A) = (1/6)/(⅓ + 0 + 1/6)

P(B3|A) = (1/6)/(1/2)

P(B3|A) = 1/3

5 0

3 years ago

The box plots show the math scores of students in two different classes. Based on the box plots, which statement is correct?

Allisa [31]

Based on the box plots, the statement which is correct is that: A. The median score of Class A is greater than the median score of Class B.

<h3>What is a box and whisker plot?</h3>

In Mathematics, a box plot is also referred to as box and whisker plot and it can be defined as a type of chart that can be used to graphically or visually represent the five-number summary of a data set with respect to locality, skewness, and spread.

Additionally, the five-number summary of any box plot (box and whisker plot) include the following:

Minimum
First quartile
Median
Third quartile
Maximum

By critically observing the box plot (box and whisker plot) which represent the math scores of students in in two different classes, we can reasonably and logically deduce the following median scores;

Median score of class A = 80

Median score of class B = 75

Therefore, a median score of 80 in Class A is greater than the median score of 75 in Class B.

Read more on box plots here: brainly.com/question/14277132

#SPJ1

6 0

1 year ago

a store has 2 boxes of markers. Each box holds 5 packs of markers. There are 8 markets in each pack. How many markers are in the

Katyanochek1 [597]

Multiply:

2 boxes x 5 packs each = 10 total packs

10 packs x 8 markers per pack = 80 total markers

Any: 80 markers

4 0

3 years ago

Jeremiah ran 5.6 miles every day for 2 weeks. How many miles did he run in all?

chubhunter [2.5K]

Answer:

78.4 miles

Step-by-step explanation:

Since there are 14 days in two weeks, you multiply 5.6 miles times 14 days to get 78.4 miles in total since 5.6 is how much he ran every DAY

3 0

3 years ago

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

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

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$