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alexira [117]
3 years ago
15

a classroom is arranged with 10 seats in the front row 12 seats in the middle row and 18 seats in the back row the teacher rando

mly assigned seats to students as they enter the classroom what is the probability of the first student who enter the classroom will be assigned a seat in the middle row?​
Mathematics
1 answer:
myrzilka [38]3 years ago
8 0

Answer:

3/10

Step-by-step explanation:

12/40 = 3/10

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postnew [5]
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(2/3)(x) = (3)(1)
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x = 3 * 3/2
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3 years ago
Can someone write down the answers pls!!!!!!!!!!
snow_lady [41]

Step-by-step explanation:

6b x>=1

6c y>7

7a x>3

7b -2<=2x<6

-1<=x<3

answer:-1,0,1,2,3

7 0
3 years ago
What are the coordinates of the pre-image of B'?
dimulka [17.4K]

Answer:

b. (-8, -2)

Step-by-step explanation:

Reflection across the x-axis

(x, y) ---> (x, -y)

B'(- 8, 2) ---> B( - 8, -2)

5 0
3 years ago
Read 2 more answers
The amount of money spent on textbooks per year for students is approximately normal.
Ostrovityanka [42]

Answer:a

a

   336.04    <  \mu < 443.96

b

  The  margin of error will increase

c

The  margin of error will decreases

d

The 99% confidence interval is  0.4107 <  p  < 0.4293

Step-by-step explanation:

From the question we are  told that

   The sample size  n =  19

    The sample mean is  \= x  = \$\  390

    The  standard deviation is  \sigma =  \$ \  120

 

Given that the confidence level is  95% then the level of significance is mathematically represented as

           \alpha = 100 -  95

          \alpha  =  5 \%

          \alpha  =  0.05

Next we obtain the critical value of \frac{\alpha }{2} from the normal distribution table

    So  

         Z_{\frac{\alpha }{2} } =  1.96

The  margin of error is mathematically represented as

      E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

=>    E = 1.96 *  \frac{120}{\sqrt{19} }

=>   E = 53.96

The 95% confidence interval is  

     \= x  -  E  <  \mu < \= x  +  E

=>   390  -   53.96   <  \mu < 390  -   53.96

=>  336.04    <  \mu < 443.96

When the confidence level increases the Z_{\frac{\alpha }{2} } also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

         n  \  \ \alpha  \ \  \frac{1}{E^2 }

So if the sample size increases the margin of error decrease

The  sample proportion is mathematically represented as

       \r p  =  \frac{210}{500}

       \r p  = 0.42

Given that the confidence level is 0.99 the level of significance is  \alpha =  0.01

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

      Z_{\frac{\alpha }{2} }  =  2.58

  Generally the margin of error is mathematically represented as

       E =  Z_{\frac{\alpha }{2} }*  \sqrt{ \frac{\r p (1- \r p )}{n} }

=>   E =  0.42 *  \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }

=>     E =  0.0093

The 99% confidence interval  is

     \r p  -  E <  p  < \r p  +  E

     0.42  -  0.0093 <  p  < 0.42  +  0.0093

     0.4107 <  p  < 0.4293

 

4 0
3 years ago
Can anyone help me and explain as well
Citrus2011 [14]

Answer:

3x2-3x+9

Step-by-step explanation:

(h-k)(3)

(X2+1-(x-2))(3)

(x2+1-x+2)(3)

3x2+3-3x+6

3x2-3x+9

6 0
3 years ago
Read 2 more answers
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