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r-ruslan [8.4K]
3 years ago
6

X 0 2 10 20

Mathematics
1 answer:
prisoha [69]3 years ago
8 0


Use two points from the graph to check if they are constant. In this case let’s use (0,50) & (2,58)

To find slope you have to do the following.

y2-y1 / x2-x1

Plug in the values, 58-50/2-0
This gives you 4.

When you use two other points, in this case (10,90) (20,130) you get a slope of 4.

This means the slope is 4.

We are trying to get an equation of y=mx+b.
We have slope already which is 4. So now we plug in y=4x + b

Now we need to find b. There is an equation to solve for b. Which is b=y1-m(x1)

But b is the y-intercept so there is already a B which is 50

So the table IS a linear equation and the equation is y=4x+50.
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Park Crest Middle School has a population of about 400 Eighth Graders.
Kobotan [32]

Answer:

10^3 times

Step-by-step explanation:

Given

Let

P \to Park Crest Middle School

T \to Total

So, we have:

P = 400

T = 406240

Required

Determine the number of times T is greater than P

This is calculated by dividing T by P.

i.e.

n = \frac{T}{P}

So, we have:

n = \frac{406240}{400}

n = 1015.6

Approximate

n \approx 1000

As a power of 10, we have:

n = 10^3

4 0
3 years ago
A simple random sample of 100 8th graders at a large suburban middle school indicated that 81% of them are involved with some ty
Sophie [7]

Answer:

The  interval is  0.7187  < p < 2.421

Step-by-step explanation:

From the question we are told that

      The  sample size is  n  = 100

       The  population  proportion is p  =  0.81

       The  confidence level is  C =  98%

The level of significance is mathematically evaluated as

     \alpha  =  100 -98

    \alpha  =  2%%

    \alpha  =  0.02

Here this level of significance represented the left and the right tail

The degree of  freedom is evaluated as

     df =  n-1

substituting value  

    df =  100 - 1

     df = 99

Since we require the critical value of one tail in order to evaluate the  98% confidence interval that estimates the proportion of them that are involved in an after school activity. we will divide the level of significance by 2

The  critical value of  \frac{\alpha}{2} and the evaluated degree of freedom is  

      t_{df , \alpha } =  t_{99 , \frac{0.02}{2}  }  = 2.33

this is obtained from the critical value table  

The standard error is mathematically evaluated as

             SE =  \sqrt{\frac{p(1-p )}{n} }  

substituting value  

           SE =  \sqrt{\frac{0.81(1-0.81 )}{100} }  

           SE = 0.0392  

The 98%  confidence interval is evaluated as

      p  - t_{df ,  \frac{\alpha }{2} } *  SE  < p <  p  + t_{df ,  \frac{\alpha }{2} }

substituting value  

     0.81  - 2.33  *  0.0392  < p <  0.81  +2.33 *  0.0392

      0.7187  < p < 2.421

     

4 0
3 years ago
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