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Allisa [31]
3 years ago
7

Find the solution to the following system using substitution or elimination:

Mathematics
1 answer:
Licemer1 [7]3 years ago
3 0

Answer:x=1

Step-by-step explanation:

y=y

y = -x + 10

y = 7x + 2

-x+10=7x+2

10=8x+2

8=8x

x=1

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In the figure above, circles A and B are tangent to each other and to the large circle. If the radius of circle A
Ne4ueva [31]

Answer:

The answer to your question is A = 200.96 u²

Step-by-step explanation:

Data

Radius A = 6

Radius B = 2

π = 3.14

Area = ?

Formula

Area = πr²

Process

1.- Calculate the radius of the large circle

Radius = radius of circle A + radius of circle B

Radius = 6 + 2

Radius = 8

2.- Substitute in the formula

Area = (3.14)(8)²

Area = 3.14(64)

Area = 200.96 u²

6 0
3 years ago
What is the answer to 3/4 divided by 9/10
meriva
The answer is <span>0.00833333333</span>
3 0
3 years ago
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Which graph shows a function with a range of all real numbers greater than or equal to –1?
cupoosta [38]

Answer:

what graph?

Step-by-step explanation: is you are going to put a quetion plz put a graph

4 0
3 years ago
Read 2 more answers
In a health club, research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard dev
Naya [18.7K]

Answer: d) 0.31

Step-by-step explanation:

Given : In a health club, research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard deviation of 5.4 minutes.

i.e. \mu=42.5 and \sigma= 5.4

It is assumed that this is a normally distributed variable.

Let x denotes the time spend by a person on treadmill.

Then, the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill.

P(30

Hence, the required probability = 0.31

Thus , the correct answer = d) 0.31

5 0
3 years ago
A consumer group is testing camp stoves. To test the heating capacity of a stove, they measure the time required to bring 2 quar
kotykmax [81]

Answer:

a

The decision rule  is  

Reject the null hypothesis

  The conclusion is  

There is sufficient evidence to show that there is a difference between the performances of these two models

b

The  95% confidence interval is  0.224   <  \mu_1 - \mu_2  < 2.776

Step-by-step explanation:

From the question we are told that

    The sample size is  n  =  36

    The first sample mean is  \= x_1   =  11.4

    The first standard deviation is  s_1 =  2.5

    The second sample mean is   \= x_2 =  9.9

     The second standard deviation is  s_2 =  3.0

      The level of significance is  \alpha  =  0.05

The null hypothesis is  H_o  :  \mu_1 - \mu_2 = 0

The alternative hypothesis is H_a :  \mu_1 - \mu_2 \ne 0

Generally the test statistics is mathematically represented as

      z =  \frac{ (\= x_1 - \= x_2 ) - (\mu_1 - \mu_2 ) }{ \sqrt{ \frac{s_1^2 }{n} + \frac{s_2^2 }{ n}  } }

=>    z =  \frac{ ( 11.4  - 9.9) - 0  }{ \sqrt{ \frac{2.5^2 }{36} + \frac{ 3^2 }{36 }  } }

=>     z = 2.3

From the z table  the area under the normal curve to the left corresponding to  2.3 is  

       P( Z >  2.3 ) =  0.010724

Generally the p-value is mathematically represented as

      p-value =  2 * P( Z >  2.3 )

=>    p-value  =  2 * 0.010724

=>    p-value  =  0.02

From the value obtained we see that  p-value  <  \alpha hence  

The decision rule  is  

Reject the null hypothesis

  The conclusion is  

There is sufficient evidence to show that there is a difference between the performances of these two models

Considering question b

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \sqrt{ \frac{s_1^2 }{n } + \frac{s_2^2}{n}}

 => E = 1.96  *    \sqrt{ \frac{2.5^2 }{ 36 } + \frac{ 3^2}{36}}

  => E = 1.276

Generally 95% confidence interval is mathematically represented as  

      ( \= x_1 - \= x_2) -E <  \mu_1 - \mu_2  < ( \= x_1 - \= x_2) + E

=>  ( 11.4 - 9.9 ) -1.276  <  \mu_1 - \mu_2 < ( 11.4 - 9.9 ) + 1.276

=>  0.224   <  \mu_1 - \mu_2  < 2.776

4 0
3 years ago
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