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UNO [17]
2 years ago
6

15 divided by 3 plus 2/3 x 7=

Mathematics
2 answers:
Maksim231197 [3]2 years ago
7 0
29/3 Because you do 15 divided by 3 +2/3 times 7 and 15 divided by 3 is 5 so then you do 5+2/3*7 which is 5+14/3 then you get 29/3,9.6
Vedmedyk [2.9K]2 years ago
5 0
I don’t know but i hope you have a good day
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3 years ago
I need help with math it’s urgent thanks
Artemon [7]

Answer:

(1, 3)

Step-by-step explanation:

 2x + 3y =  11      ------------(i)

-4x + 2y = 2        ------------(ii)

Multiply equation (i) by 2.

(i)*2              4x + 6y = 22

(ii)               <u>-4x  + 2y = 2</u>                  {Now add and x will be eliminated}

                            8y = 24

y= 24/8

y = 3

Plugin the value of y in equation (i)

2x + 3*3 = 11

2x + 9 = 11

2x = 11-9

2x = 2

x = 2/2

x = 1

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3 years ago
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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
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