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2 years ago

Find the area of the figure

Mathematics

2 answers:

Kitty [74]2 years ago

8 0

Answer:

38units²

Step-by-step explanation:

10*3=30 + 4*2=8

got 4 cuz 6+x=10 then x=4

got 2 cuz 5-3=x so x=2

Yuliya22 [10]2 years ago

3 0

Answer:

the area would be, 38 feet.

Step-by-step explanation:

This is because you just need to either draw or think that there is a line continuing the side length of 6' to create 2 rectangles. there would be one 3x10' rectangle and on 4x2'. that adds up to 38'

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J^2 over 2j times 2j over 3g

Serhud [2]

Answer:

Step-by-step explanation:

given $\frac{j^2}{2j}$ × $\frac{2j}{3g}$

the 2j on the denominator of the first fraction cancels with the 2j on the numerator of the second fraction, leaving the expression in simplified form as

= $\frac{j^2}{3g}$ → D

5 0

2 years ago

What does jury duty mean?

JulijaS [17]

Jury duty is where you have to server your duty being on a jury for a court case. On a jury duty, you vote if you think the person on trial is guilty or not.

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2 years ago

What is 7+10x-2x-2=-3

Ivahew [28]

7+10x-2x-2=-3

—-Combine like terms...10x-2x=8x and 7-2=5

8x+5=-3
——subtract 5 on both sides.. the 5’s will cancel each other and -3-5=-8

8x=-8
——Divide 8 on both sides...the 8’s will cancel out and -8 divided by 8 = -1

Answer: x = -1

6 0

2 years ago

A boat travels 320 miles in 11 hours (with a constant speed), How much

Arada [10]

Answer: 4.2625

Step-by-step explanation:

11/320=x/124

11*124=320x

1364=320x

x=4.2625

Hopefully I was helpful!

3 0

2 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from to , the response time for medical aid var

-BARSIC- [3]

Answer:

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

,10.3

,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

3 0

2 years ago