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Arte-miy333 [17]
4 years ago
6

Can somebody tell me the answer and maybe tell me how you got it too.​

Mathematics
1 answer:
asambeis [7]4 years ago
7 0

To get t on one side, we have to use inverse operations to get rid of the 5.

1. Use inverse operation of multiplication to get rid of 5. Which would be division.

2. Divide both sides by 5.

5t/5 = 7 1/2 / 5

3. Solve.

5t/5 = t and 7 1/2 /5 = 1 1/2

7.5 / 5 = 1.5

Therefore, your answer would be C. divide both sides by 5, and the answer is 1 1/2.

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The number of students in a chess club decreased from 17 to 10. What is
Svetllana [295]

Answer:

  • 41.18%

Step-by-step explanation:

  • Initial number = 17
  • Final number = 10

<u>Decrease in number:</u>

  • 17 - 10 = 7

<u>Percent decrease:</u>

  • 7/17*100% = 41.18% (rounded)
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2 years ago
Write the equation of the line in fully simplified slope-intercept form.
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Y = -6x + 3

The slope is -6 because you move down 6 units and over 1, and -6/1 = -6. The y-intercept is 3 because the line touches the y-axis at (0,3).
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What is 0.987 tenths?
almond37 [142]

Answer:

0.0987

Step-by-step explanation:

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3 years ago
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
4 years ago
39,892 to the nearest 1000 on the number line
Iteru [2.4K]

I'm pretty sure it would be 40,000 because if i'm not mistaken you meant the nearest 10,000 since the number is 39,892.

7 0
3 years ago
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