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3 years ago

Michelle went shopping for some clothes and found a pair of slacks that cost

Mathematics

1 answer:

34kurt3 years ago

8 0

Answer:

The sirt cost $7.5 before tax.

Step-by-step explanation:

Since the pair of slacks cost twice as much as the shirt, we have:

slacks = 2*shirt

And the total purchase was $22.5, so we have:

slacks + shirt = 22.5

We can create a system of equations such as:

slacks = 2*shirt

slacks + shirt = 22.5

If we apply the first formula on the second we can solve for the value of the shirt, we have:

2*shirt + shirt = 22.5

3*shirt = 22.5

shirt = 22.5/3 = 7.5

So the slacks cost:

slacks = 2*7.5 = 15

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Step-by-step explanation:

__________________________________________________________

FACTS TO KNOW BEFORE SOLVING :-

$a^x \times a^y = a^{x+y}$

In an equation , if the bases are same in both L.H.S. & R.H.S. then , the power of the bases on both the sides of equation should be equal. For e.g. : $a^x = a^y$ ⇒ $x = y$ [∵ Bases are equal on both the sides]

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$4 \times 8^{2x+1} = 32^{x+2}$

Lets express it in terms of 2.

$=> 2^2 \times 2^{3 (2x+1)} = 2^{5(x+2)}$

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A)Mean :10.65

Median =10.7

Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}$

Mean = 10.65

Median: The mid value of the data

Data in ascending order

8.3

10.3

10.5

10.6

10.7

10.9

11.2

11.5

11.8

n=10(even)

$Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7$

Mode : the most occurring frequency

10.7 is occurring twice while others are occurring once

So, Mode is 10.7

B) Range = Maximum - Minimum=11.8-8.3=3.5

Standard deviation : $\sqrt{\frac{\sum(x-\bar{x})^2}{n}}$

Standard deviation : $\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}$

Standard deviation :0.89916

8.3

,10.3

,10.5

,10.6

,10.7

,10.9

,11.2

,11.5

,11.8

For Q1 ( Median of lower quartile )

8.3

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,10.5

,10.6

,10.7

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5$

For Q3( Median of Upper quartile )

10.7

,10.9

,11.2

,11.5

,11.8

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2$

IQR = Q3-Q1=11.2-10.5=0.7

Range :(Q1-1.5IQR, Q3-1.5IQR)

Range : $(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)$

Range :(9.45, 10.15)

8.3 does not lie in this interval

So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times

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