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Sphinxa [80]
3 years ago
13

3/7 of students at a school are boys . If there are 2282 students at the school. How many are girls .

Mathematics
1 answer:
Natasha2012 [34]3 years ago
7 0
If 3/7 are boys, then 4/7 have to be girls.

2282 / 7 = 326
326 x 4 = 1304

There are 1304 girls. 
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dimaraw [331]
You add 2x and 7x to get 9x
so the answer is 9x+5y
6 0
3 years ago
The value of 3 in 6300 is how many times the value of 3 in 530
deff fn [24]
The value of the 3 in 6300 os 300 and it is ten times as many as the 3 in 530
8 0
3 years ago
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(100 points) Give a step by step explanation on how you solved this problem
Lina20 [59]
Use substitution
Given both y values, you know that they have to be equal

3x + 3 = x - 1
2x + 3 = -1
2x = -4, divide by 2
x = -2
Now plug x value into any of the equation to find y value
y = 3(-2) + 3
y = -6 + 3, y = -3

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4 0
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
Use the formula for compound amount:$14,800 at 6% compounded semiannually for 4 years
GalinKa [24]

SOLUTION

Given the question in the question tab, the following are the solution steps to answer the question.

STEP 1: Write the formula for calculating compound amount

A=P(1+\frac{r}{n})^{nt}

where

A = final compounded amount

P = initial principal balance

r = interest rate

n = number of times interest applied per time period

t = number of time periods elapsed

STEP 2: Write the given data

Semiannually means that n will be 2

P=14,800,r=\frac{6}{100}=0.06,n=2,t=4

STEP 3: Calculate the compound amount

\begin{gathered} A=14800(1+\frac{0.06}{2})^{2\times4}\Rightarrow A=14800(1+0.03)^{2\times4} \\ A=14800(1.03)^8 \\ A=14800\times1.266770081 \\ A=\text{\$}18,748.1972 \end{gathered}

Hence, the compounded amount after 4 years is $18,748.1972

6 0
1 year ago
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