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dsp73
3 years ago
11

Lauren simplified the expression m^-7 -5 as shown

Mathematics
2 answers:
morpeh [17]3 years ago
7 0
The answer is B cause you multiplied them not add or subtract 
melamori03 [73]3 years ago
4 0

Answer:

b

Step-by-step explanation:

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(25 POINTS) Write the inequality of the following graph?
Ainat [17]

Answer:

f(x) > -x + 2

Step-by-step explanation:

First I found the equation of the line.

The y intercept is at 2 so I know to add 2 at the end

The slope is -1 since it is going downwards

Since the top side is shaded, that represents greater than, if it was shaded downwards, it would be less than

Putting all that together, the equation is f(x) > -x + 2

5 0
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
You need a loan for $1000, and you can take this loan out with Company A or Company B. Company A charges an 8% interest rate and
Wittaler [7]

Answer: Company b

Step-by-step explanation:

because company B is only 5percent it is better and can pay in 8 years

5 0
3 years ago
Find the volume of the described solid s
Serga [27]

Answer:lol it’s 697

Step-by-step explanation: it could be

8 0
3 years ago
Please help asap will mark brainliest
melisa1 [442]
The second one because since n is not negative then nothing will be in the denominator
7 0
3 years ago
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