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grin007 [14]
3 years ago
12

3x+2y=10 12x+5y=25 Solve by substitution or elimination

Mathematics
1 answer:
zaharov [31]3 years ago
5 0
3x + 2y = 10 ....multiply by -4
12x + 5y = 25
-------------------
-12x - 8y = -40 (result of multiplying by -4)
12x + 5y = 25
------------------add
-3y = - 15
y = -15/-3
y = 5

3x + 2y = 10
3x + 2(5) = 10
3x + 10 = 10
3x = 10 - 10
3x = 0
x = 0

solution is (0,5)
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1. 10^2

2. 10^5

3. 10^4

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3 years ago
What is the volume of this rectangular prism?<br> 10/3 <br> 4/5<br> 1/5
Lunna [17]

Answer:

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Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0
3 years ago
A shopkeeper allows 10% discount on the marked price of a bicycle. If a costomer pays Rs 4068 with 13% VAT find the marked price
Over [174]

\large{ \tt{❁ \: S \: O \: L \: U \: T \: I \: O \: N \: ❁}}

  • We're provided - Discount % = 10 % , Cost with VAT [ SP with VAT ] = Rs 4068 & VAT % = 13%. We're asked to find out the marked price of the bicycle. Let's start :

\large{ \tt{❇ \: FIND \: SP \: WITHOUT \: VAT \: / \: SP\: ❇}}

\large{ \tt{❃ \: SP  \: with \: VAT= SP + VAT\% \: of \: SP}}

\large{ \tt{⇢ \: 4068 = SP + 13\% \: of \: SP}}

\large{ \tt{⇢ \: 4068 = SP +  \frac{13}{100}  \: sp}}

\large{ \tt{⇢ \: 4068 =  \frac{100 \:  \: SP + 13 \: SP}{100} }}

\large{ \tt{⇢ \: 4068 =  \frac{113 \: SP}{100} }}

\large{ \tt{⇢ \: 113 \: SP = 406800}}

\large{ \tt{⇢ \: SP =  \frac{406800}{113} }}

\large{ \tt{⇢ \: SP = 3600}}

  • Hence , SP = Rs 3600

\large{ \tt{✽ \: NOW , \: FIND \: THE \: MP \:✽ }}

  • Let Marked Price [ MP ] be x.

\large {\tt{❃ \: SP = MP - dis\% \: of \: MP}}

\large{ \tt{⇾ \: 3600 = x - 10\% \: of \: x}}

\large{ \tt{⇾ \: 3600 = x -  \frac{10}{100} } \: x}

\large{ \tt{⇾ \: 3600 =  \frac{100x - 10x}{100} }}

\large{ \tt{⇾ \: 3600 =  \frac{90x}{100} }}

\large{ \tt{⇾ \: 90x = 360000}}

\large{ \tt{⇾ \: x =  \frac{360000}{90} }}

\large{ \tt{⇾ \: x = Rs \: 4000}}

\large{ \boxed{ \boxed{ \tt{☂ \: OUR \: FINAL \: ANSWER :  \boxed {\tt {\: Rs \: 4000}}}}}}

  • Hope I helped! Let me know if you have any questions regarding my answer and don't hesitate to reach out to me if you need any assistance! :)
6 0
3 years ago
Among 35 students in a class, 20 of them earned A's on the midterm exam, 13 earned A's on the nal exam, and 12 did not earn A's
enyata [817]

Answer:

The correct answer is "\frac{2}{7}".

Step-by-step explanation:

According to the question,

Number of students,

= 35

A ! mid term A,

n(A) = 20

B : final A,

n(B) = 13

Didn't start,

n(A^c \cap B^c) = 12

Now,

⇒ n(A \cup B)=35-12

                   =23

then,

⇒ n(A \cup B)=20+13-23

                   =10

hence,

The probability will be:

⇒ P(A \cap B)=\frac{10}{35}

⇒                 =\frac{2}{7}

7 0
3 years ago
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