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jeka57 [31]
3 years ago
14

Nicholas says that (2^6)(2^6) equals 2^12, which also equals 4^6. is this true? PLEASE EXPLAIN

Mathematics
2 answers:
ahrayia [7]3 years ago
7 0

a^b \cdot a^c= a^{b+c}\\ a^c \cdot b^c=(ab)^c\\\\ 2^6 \cdot 2^6 =2^{6+6}=\underline{2^{12}}=(2\cdot2)^6=\underline{4^6}

So, yes, it's true.

Umnica [9.8K]3 years ago
5 0

When you multiply two numbers in exponential notation whose bases are the same, you just keep the base the same and add the exponents.

In this case, you have (2^6) * (2^6) = 2^(6 + 6) = 2^12.

We can change the base of 4^6 into 2:

(2^2)^6

Is this the same as 2^12? Well, when we have an expression like that above, we multiply the exponents together, so we get: (2^2)^6 = 2^12.

Therefore, Nicholas is correct.

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A random sample of 20 items is selected from a population. When computing a confidence interval for the population mean, what nu
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Answer:

For this case since the sample size is lower than 30 , n =20<30, is not appropiate use the normal standard distribution to calculate the confidence interval for the mean, and then for this case is better use the t distribution since takes in count the correction factor to approximate the distribution of the true parameter.

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

For this case since the sample size is lower than 30 , n =20<30, is not appropiate use the normal standard distribution to calculate the confidence interval for the mean, and then for this case is better use the t distribution since takes in count the correction factor to approximate the distribution of the true parameter.

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

8 0
4 years ago
Help asap show work
katovenus [111]

Answer:  see below

<u>Step-by-step explanation:</u>

17) 5n - 3 > -2n + 4     and    9 - 4n ≥ -2n + 1

   <u> +2n   </u>    <u> +2n      </u>              <u>    +2n </u>   <u>+2n      </u>

   7n - 3   >     4                    9 - 2n  ≥       1

        +3         +3                  -9                 -9

    7n       >      7                       -2n   ≥     -8

  <u> ÷7     </u>    <u>     ÷7  </u>                   <u> ÷ -2 </u>  ↓  <u> ÷ -2 </u>

          n  >     1         and            n   ≤   4                  

Graph:   o----------------- ·

              1                     4

18) 9k - 2 ≤ 9 + 10k < 9 - 4k

     9k - 2 ≤ 9 + 10k      and     9 + 10k < 9 - 4k

  <u> -10k     </u>    <u>     -10k </u>               <u>       +4k </u>   <u>    +4k  </u>

     -k - 2 ≤ 9                             9 + 14k < 9

   <u>      +2</u>   <u>+2         </u>                 <u> -9          </u>  <u>-9         </u>

     -k      ≤  11                                  14k  <  0

 <u> ÷ -1    </u>   ↓<u> ÷ -1  </u>                      <u>       ÷14 </u>    <u>÷14      </u>

          k  ≥ -11                and               k < 0

Graph:    · -----------------o

              -11                  0    

19) 4 - n ≤ 10 + 5n ≤ 6n + 1

     4 - n ≤ 10 + 5n       and     10 + 5n ≤  6n + 1

   <u>    -5n </u>   <u>      -5n  </u>                 <u>      -6n </u>   <u>-6n     </u>

   4 - 6n ≤ 10                           10  -  n  ≤       1    

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        -6n ≤ 6                                    -n ≤   -9

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           n ≥ -1              and                 n ≥ 9

Graph:    · -----------------→

              -9                  

20) 4p - 5 ≤ 2p + 9 ≤ 4p + 7

     4p - 5 ≤ 2p + 9     and     2p + 9 ≤ 4p + 7

  <u> -2p       </u>   <u>-2p    </u>                 <u> -4p    </u>   <u>-4p      </u>

    2p - 5 ≤         9                -2p + 9  ≤        7    

    <u>      +5   </u>  <u>    +5    </u>             <u>        -9  </u>    <u>     -9  </u>

       2p  ≤       14                   -2p       ≤      -2

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         p ≤       7          and         p    ≥      1

Graph:    · ----------------- ·

              1                    7

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