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Aneli [31]
3 years ago
14

Which is the best first step when solving the following system of equations?

Mathematics
2 answers:
exis [7]3 years ago
8 0
B. Substitute 9x for y in the first equation
pychu [463]3 years ago
5 0

Answer:

IT B

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B2m%20-%201%7D%20%20-%20%20%5Cfrac%7B1%7D%7Bm%20%2B%202%7D%20%20%3D%201%20%
Scilla [17]

Answer:

m = 1

Step-by-step explanation:

<u>2(m+2) - (2m-1)</u> = <u>5</u>

(m+2)(2m-1)          3

2m² -1m + 4m -2 = 3

2m² + 3m - 5 = 0

(2m+5)(m-1) = 0

m = 1    This is the only value that satisfies the equation

m = -5/2

3 0
3 years ago
The length of a shadow of a building is 29m. The distance from the top of the building to the tip of the shadow is 35m. Find the
TEA [102]

Answer:

19.5m

Step-by-step explanation:

Use the Pythagorean theorem.

a^2 + b^2 = c^2

The longest side, 35m, is c and the base of the triangle, 29m, is b.

a^2 + 29^2 = 35^2

Simplify.

a^2 + 841 = 1225

Get a by itself by adding -841 to both sides to cancel out the 841.

a^2 = 384

Find the square root of both sides to get rid of the exponent.

square root of 384 = ~19.5959

a = 19.5

3 0
4 years ago
How do you write in words 0.102
Svet_ta [14]

point one zero two

or

zero and one zero two thousandths

6 0
3 years ago
A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are list
alexandr402 [8]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

Yes the students are reasonably good at estimating one​ minute

a

  Option A is  correct

b

  The test statistics is  t  =  0.354

Step-by-step explanation:

From the question we are told that

    The  set of data is  

               68, 82 ,  38 ,  62 , 41, 25 , 57 ,  64, 67, 47, 61, 71, 91, 87, 64

     The  population mean is  \mu  = 60

The  level of significance is given as\alpha =  0.01

    The  critical value for this level of significance obtained from the normal distribution table is  

         Z_{\alpha } =  2.33

   The  null hypothesis is  

           H_o  :  \mu  =  60 \ seconds

   The alternative hypothesis is  

          Ha :  \mu \ne 60 \ seconds      

Generally the sample mean is mathematically represented as

       \= x  =  \frac{\sum x_i }{n  }

where  n = 15

 So  

      \= x  =  \frac{ 68+ 82 +  38 +  62 + 41+ 25 + 57 + 64+67+ 47+ 61+ 71+ 91+ 87+ 64}{15}

      \= x  =  61.67

The standard deviation is mathematically represented as

     \sigma  =\sqrt{  \frac{ \sum  (x_i -  \= x )^2}{n} }

substituting values

      \sigma  =\sqrt{  \frac{ \sum  (68-61.67)^2 + ( 82-61.67 )^2  +   (38 -61.67)^2 +  62-61.67)^2 + (41-61.67)^2 +(25-61.67)^2 +( 57+61.67)^2 }{15} }           \sqrt{  \frac{  ( \cdot  \cdot  + (  64-61.67)^2 + (67-61.67)^2 +(47-61.67)^2 + (61-61.67)^2+  (71-61.67)^2 + (91 -61.67)^2+( 87-61.67)^2 +  (64-61.67)^2}{15} }=>   \sigma  =  18.23

The  test statistics is evaluated as  

      t  =  \frac{\= x  - \mu  }{ \frac{\sigma }{\sqrt{n} } }

substituting values

       t  =  \frac{61.67  -60  }{ \frac{18.23 }{\sqrt{ 15} } }

      t  =  0.354

Now comparing the statistics and the critical value of the level of significance we see that the the test statistics is less than the critical value

Hence the we fail to reject the null hypothesis which mean that these times are from a population with a mean equal to 60 seconds

So we can state that yes the students are reasonably good at estimating one minute given that the sample mean is not far from the population mean

4 0
3 years ago
16 24 36 54 COMMON RATIO
Yakvenalex [24]

All of them can be divided by 2 so ratio 1:2 is a common ratio

4 0
3 years ago
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