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nikitadnepr [17]
3 years ago
15

What is the answer to 9x8 2/3

Mathematics
2 answers:
Ira Lisetskai [31]3 years ago
7 0
The answer is 78. This is because you change 8 2/3 into a improper fraction which is 26/3. Then do 26/3×9/1. You should cross cancel so cancel out 3 and change it into 1 and change 9 into 3. So now your problem is 26×3=78.
~JZ
Hope it helps.
Mila [183]3 years ago
6 0
9 x 8 2/3.

First, make 9 a fraction: 9=9/1.

Then, make 8 2/3 a improper fraction: 8 2/3=26/3.

Now, we multiply: 9/1×26/3=234/3.

Finnally, we simplify: 234/3=78.

Therefore, the answer is 78.
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(3 pt) Marie has $830 in her bank account and withdraws $60 each month. Denise has $970 in her bank account and withdraws $80 ea
Lady bird [3.3K]
The answer is c because if you do 7x60 you'll get 420 and 830-420=410 and 7x80 you'll get 560 and 970-560 is also 410.
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3 years ago
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Which of the following graphs shows the solution set for the inequality below? 3|x + 1| < 9
Bas_tet [7]

Step-by-step explanation:

The absolute value function is a well known piecewise function (a function defined by multiple subfunctions) that is described mathematically as

                                 f(x) \ = \ |x| \ = \ \left\{\left\begin{array}{ccc}x, \ \text{if} \ x \ \geq \ 0 \\ \\ -x, \ \text{if} \ x \ < \ 0\end{array}\right\}.

This definition of the absolute function can be explained geometrically to be similar to the straight line   \textbf{\textit{y}} \ = \ \textbf{\textit{x}}  , however, when the value of x is negative, the range of the function remains positive. In other words, the segment of the line  \textbf{\textit{y}} \ = \ \textbf{\textit{x}}  where \textbf{\textit{x}} \ < \ 0 (shown as the orange dotted line), the segment of the line is reflected across the <em>x</em>-axis.

First, we simplify the expression.

                                             3\left|x \ + \ 1 \right| \ < \ 9 \\ \\ \\\-\hspace{0.2cm} \left|x \ + \ 1 \right| \ < \ 3.

We, now, can simply visualise the straight line,  y \ = \ x \ + \ 1 , as a line having its y-intercept at the point  (0, \ 1) and its <em>x</em>-intercept at the point (-1, \ 0). Then, imagine that the segment of the line where x \ < \ 0 to be reflected along the <em>x</em>-axis, and you get the graph of the absolute function y \ = \ \left|x \ + \ 1 \right|.

Consider the inequality

                                                    \left|x \ + \ 1 \right| \ < \ 3,

this statement can actually be conceptualise as the question

            ``\text{For what \textbf{values of \textit{x}} will the absolute function \textbf{be less than 3}}".

Algebraically, we can solve this inequality by breaking the function into two different subfunctions (according to the definition above).

  • Case 1 (when x \ \geq \ 0)

                                                x \ + \ 1 \ < \ 3 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 3 \ - \ 1 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 2

  • Case 2 (when x \ < \ 0)

                                            -(x \ + \ 1) \ < \ 3 \\ \\ \\ \-\hspace{0.15cm} -x \ - \ 1 \ < \ 3 \\ \\ \\ \-\hspace{1cm} -x \ < \ 3 \ + \ 1 \\ \\ \\ \-\hspace{1cm} -x \ < \ 4 \\ \\ \\ \-\hspace{1.5cm} x \ > \ -4

           *remember to flip the inequality sign when multiplying or dividing by

            negative numbers on both sides of the statement.

Therefore, the values of <em>x</em> that satisfy this inequality lie within the interval

                                                     -4 \ < \ x \ < \ 2.

Similarly, on the real number line, the interval is shown below.

The use of open circles (as in the graph) indicates that the interval highlighted on the number line does not include its boundary value (-4 and 2) since the inequality is expressed as "less than", but not "less than or equal to". Contrastingly, close circles (circles that are coloured) show the inclusivity of the boundary values of the inequality.

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2 years ago
A total of 150 students have taken an Algebra 2 final exam. The scores are normally distributed with a mean of 71% and standard
WARRIOR [948]

Answer:

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Step-by-step explanation:

Note that 65% and 71% are both 1 standard deviation from the mean (71%).  According to the empirical rule, 68% of scores lie within 1 std. dev. of the mean.

68% of 150 students would be 0.68(150 students) = 102 students

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Please go answer my recent questions ASAP.
Fiesta28 [93]

Answer:

OK, Im going to help you!!!!!!!!!!!

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The image of the point (-5,4) under a reflection across the Y-axis is (5,4). True or false
Vanyuwa [196]

The answer is true. A reflection is a translation over an axis

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