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siniylev [52]
2 years ago
6

10) in a school there are 1500 students. if 60% are boys, find the numbers of boys.​

Mathematics
2 answers:
Aleks04 [339]2 years ago
6 0

Answer:

there are 900 boys in the school

15/100*1500

Debora [2.8K]2 years ago
4 0

Answer:

900

Step-by-step explanation:

.6 * 1500

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I need to know how to solve this problem because I’m very confused and it doesn’t explain what it is or how to do it in my textb
Lostsunrise [7]

Answer:

x1 = -4; x2 = 10

Step-by-step explanation:

1) Expand the module as two separate equations:

x - 3 = 7

x - 3 = -7

2) Solve the equations:

x = 10

x = -4

=> x1 = -4; x2 = 10

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Evaluating expressions for the value of x 4.5x-12 x=2
Svetradugi [14.3K]

X = 2

4.5(2) -12

9 -12

= -3

Answer is -3

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A function is defined by f(x)=x/4, what is f(-8)
barxatty [35]
F(x) = x/4
so
f(-8) = -8/4 = -2
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f(-8) = -2
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3 years ago
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During her eight-hour shift, Sami vacuumed 104 cars. How many cars can Sami vacuum per hour?
Nadusha1986 [10]
104/ 8 = 13.

Every hour she vacuumed 13 cars.
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3 years ago
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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:

A)

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

10.7

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

C)

8.3

,10.3

,10.5

,10.6

,10.7

,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
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