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Lynna [10]
2 years ago
9

Write the next whole number after 4444 in the base-five system.

Mathematics
1 answer:
svet-max [94.6K]2 years ago
7 0
<h3>Answer:   10000 in base 5</h3>

====================================================

Explanation:

4+1 = 5 in base 10

But in base 5, the digit "5" does not exist.

The only digits in base five are: 0, 1, 2, 3, 4

This is similar to how in base ten, the digits span from 0 to 9 with the digit "10" not being a thing (rather it's the combination of the digits "1" and "0" put together).

----------------

Anyways let's go back to base 5.

Instead of writing 4+1 = 5, we'd write 4+1 = 10 in base 5. The first digit rolls back to a 0 and we involve a second digit of 1.

Think how 9+1 = 10 in base 10.

Similarly,

44+1 = 100 in base 5

444+1 = 1000 in base 5

4444+1 = 10000 in base 5

and so on.

----------------

Here are the first few numbers in base 5, when counting up by 1 each time.

0, 1, 2, 3, 4,

10, 11, 12, 13, 14,

20, 21, 22, 23, 24,

30, 31, 32, 33, 34,

40, 41, 42, 43, 44,

100, 101, 102, 103, ...

Notice each new row is when the pattern changes from what someone would expect in base 10. This is solely because the digit "5" isn't available in base 5.

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Answer:

AC = 23 so it be D

Step-by-step explanation:

plug AC = DF ---> 21x-19 = 3x+17

sub. 3x by both sides --> 18x-19 = 17

add 19 to both sides --> 18x = 36

divide both sides by 18, x = 2

plug x back into 21x-19 to find AC

21(2) - 19

42 - 19 = 23

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4 years ago
Solid fats are more likely to raise blood cholesterol levels than liquid fats. Suppose a nutritionist analyzed the percentage of
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Answer:

t = 31.29

Step-by-step explanation:

Given

\begin{array}{ccccccc}{Stick} & {25.8} & {26.9} & {26.2} & {25.3} & {26.7}& {26.1} \ \\ {Liquid} & {16.9} & {17.4} & {16.8} & {16.2} & {17.3}& {16.8} \ \end{array}

Required

Determine the test statistic

Let the dataset of stick be A and Liquid be B.

We start by calculating the mean of each dataset;

\bar x =\frac{\sum x}{n}

n, in both datasets in 6

For A

\bar x_A =\frac{25.8+26.9+26.2+25.3+26.7+26.1}{6}

\bar x_A =\frac{157}{6}

\bar x_A =26.17

For B

\bar x_B =\frac{16.9+17.4+16.8+16.2+17.3+16.8}{6}

\bar x_B =\frac{101.4}{6}

\bar x_B =16.9

Next, calculate the sample standard deviation

This is calculated using:

s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}

For A

s_A = \sqrt{\frac{\sum(x - \bar x_A)^2}{n-1}}

s_A = \sqrt{\frac{(25.8-26.17)^2+(26.9-26.17)^2+(26.2-26.17)^2+(25.3-26.17)^2+(26.7-26.17)^2+(26.1-26.17)^2}{6-1}}

s_A = \sqrt{\frac{1.7134}{5}}

s_A = \sqrt{0.34268}

s_A = 0.5854  

For B

s_B = \sqrt{\frac{\sum(x - \bar x_B)^2}{n-1}}

s_B = \sqrt{\frac{(16.9 - 16.9)^2+(17.4- 16.9)^2+(16.8- 16.9)^2+(16.2- 16.9)^2+(17.3- 16.9)^2+(16.8- 16.9)^2}{6-1}}

s_B = \sqrt{\frac{0.92}{5}}

s_B = \sqrt{0.184}

s_B = 0.4290

Calculate the pooled variance

S_p^2 = \frac{(n_A - 1)*s_A^2 + (n_B - 1)*s_B^2}{(n_A+n_B-2)}

S_p^2 = \frac{(6 - 1)*0.5854^2 + (6 - 1)*0.4290^2}{(6+6-2)}

S_p^2 = \frac{2.6336708}{10}

S_p^2 = 0.2634

Lastly, calculate the test statistic using:

t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_B)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}

We set

\mu_A = \mu_B

So, we have:

t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_A)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}

t = \frac{(\bar x_A - \bar x_B) }{\sqrt{S_p^2/n_A +S_p^2/n_B}}

The equation becomes

t = \frac{(26.17 - 16.9) }{\sqrt{0.2634/6 +0.2634/6}}

t = \frac{9.27}{\sqrt{0.0878}}

t = \frac{9.27}{0.2963}

t = 31.29

<em>The test statistic is 31.29</em>

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\frac{x}{(x+2)(x+1)}-\frac{1}{(x+2)(x+1)}

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Subtract the numerators and write one denominator;

\frac{x-1}{(x+2)(x+1)}}

Rewrite to get;

\frac{x-1}{x^2+3x+2}}

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