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valentinak56 [21]
3 years ago
13

Please help me please really need help please

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
4 0
The answer is Y = 2/3x - 4

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Round 5,939576 to the nearest hundred thousand
TiliK225 [7]

Answer:

5,939,576.000

Step-by-step explanation:

none added explanation, hoped it help!

7 0
4 years ago
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What is the 6th term of the geometric sequence f(1) = 12, f(n) = -2/3f(n-1)
Troyanec [42]
\bf \begin{array}{ccll}
\stackrel{n^{th}}{term}&value\\\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\\\
1&12\\\\
2&-\frac{2}{3}f(2-1)\\\\
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&-\frac{2}{3}(12)\\\\
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3&-\frac{2}{3}f(3-1)\\\\
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&-\frac{2}{3}(-8)\\\\
&\frac{16}{3}
\end{array}

\bf \begin{array}{cccl}
\qquad &\qquad \\
4&-\frac{2}{3}f(4-1)\\\\
&-\frac{2}{3}f(3)\\\\
&-\frac{2}{3}\left( \frac{16}{3} \right)\\\\
&-\frac{32}{9}\\\\
5&-\frac{2}{3}f(5-1)\\\\
&-\frac{2}{3}f(4)\\\\
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5 0
3 years ago
0.08 is 10 times as great as
Wittaler [7]
0.08/10= 0.008. 0.08 is ten times as great as 0.008
7 0
3 years ago
Help please? And show work?
In-s [12.5K]
Answer the answer is 25




Explanation:
6 0
3 years ago
following frequency distribution shows the daily expenditure on milk of 30 households in a locality:Daily expenditure on milk(in
valina [46]

<em>Note:</em> <em>As you missed to identify what we have to find in this question. But, after a little research, I am able to find that we had to find the Mode for the data given in your question. So, I am assuming we have to calculate the the Mode. Hopefully, it would clear your concept regarding this topic.</em>

Answer:

The mode of the data = 75

Step-by-step explanation:

Lets visualize the given data in a table to show the frequency distribution:

<em>Daily expenditure on milk (in Rs)           Number of households</em>

<em>                  0-30                                                           5</em>

<em>                 30-60                                                          6        </em>

<em>                 60-90                                                          9</em>

<em>                 90-120                                                         6</em>

<em>                 120-150                                                        4</em>

Here the maximum frequency is 9.

So, modal class is <em>60-90.</em>

<em />

As the formula to calculate the mode:

Mode = l_{1} + h (\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} )

Here, the maximum

l_{1} =60, f_{1} =9, f_{0}=6, f_{0}=6, h=30

l= is the lower limit of the class

f_{1} = is the frequency of the modal class

f_{0} = is the frequency of the previous modal class

f_{2} = is the frequency of the next previous modal class

l= is the class size

So,

Mode = l_{1} + h (\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} )

Mode = 60 + 30 (\frac{9-6}{2(9)-6-6} )

Mode = 60 + \frac{(30)(3)}{6}

Mode = 60 + 15=75

∴ The mode of the data = 75

<em>Keywords: mode, frequency distribution</em>

<em> Learn more about mode and frequency distribution from brainly.com/question/14354368</em>

<em> #learnwithBrainly </em>

5 0
3 years ago
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