All categories

2 years ago

May I ples have a little help pless Simplify using the laws of exponents [2^4/2^ 3]^5

Mathematics

1 answer:

xxMikexx [17]2 years ago

5 0

$\left[\dfrac{2^4}{2^3} \right]^5 \\\\\\=[2^{4-3}]^5\\\\=[2^1]^5\\\\=2^5\\\\=32$

You might be interested in

How can i solve for y in this problem? 2x-y=3

pshichka [43]

Hello from MrBillDoesMath!

Answer:

y = 2x - 3

Discussion:

2x - y = 3 => add "y" to both sides

2x - y + y = 3 + y => as -y+y = 0

2x = 3 + y => subtract 3 from each side of the equation

2x-3 = y

Thank you,

MrB

4 0

3 years ago

If a population grows by 69%, what is the multiplier?

Alexeev081 [22]

The percentage duhhhh

7 0

2 years ago

The ages of the Oscar-winners for best actor and actress for the years 1970 through 1990 are given below: Women: 34, 34,26,37,42

Artemon [7]

Answer:

See explanation below.

Step-by-step explanation:

Numerical way

We can calculate the mean and the standard deviation from a sample with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s= \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

We have the following dataset for women: 34,34,26,37,42,41,35,31,41,33,30,74,33,49,38,61,21,41,26,80,72

And the mean and deviation are:

$\bar X_F = 41.857$

$s_F = 16.399$

We can calculate the median ordering the data like this:

21,26,26,30,31,33,33,34,34,35,37,38,41,41,41,42,49,61,72,74,80

And the median would be on the 11 position of the data ordered, for this case is:

$median_F = 37$

$Minimum_F = 21 , Maximum_F = 80$

And the range is given by: $Range_F = 80-21=59$

We can also calculate the coefficient of variation given by:

$\hat{CV} =\frac{s}{\bar X} = \frac{16.399}{41.857}=0.392$

We have the following dataset for men:

43, 40,48,48,56,38,60,32,40,42,37,76,39,55,45,35,61,33,51,32,42

And the mean and deviation are:

$\bar X_M = 45.381$

$s_M = 11.218$

We can calculate the median ordering the data like this:

32, 32,33,35,37,38,39,40,40,42,42,43,45,48,48,51,55,56,60,61,76

And the median would be on the 11 position of the data ordered, for this case is:

$median_M = 42$

$Minimum_M = 32 , Maximum_F = 76$

And the range is given by: $Range_M = 76-32=44$

We can also calculate the coefficient of variation given by:

$\hat{CV} =\frac{s}{\bar X} = \frac{11.218}{45.381}=0.247$

We se that the mean for the male groups is higher than the mean for female group. The deviation is higher for the female group comapred to the male group. We have more variation for the female group as we can see on the range and the coeffcient of variation.

Graph

We can see the histograms for each group on the figures attached.

As we can see the histogram for males is skewed to the right with the median lower than the mean.

By the other hand the histogram for the female group seems to be skewed to the right also but with missing values on the interval 50-60, for this reason we have more variation compared to the male group.