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mash [69]
3 years ago
14

HELP PLEASEEE !!!!!!!!!!!!

Mathematics
2 answers:
Fantom [35]3 years ago
8 0
Multiply 15000 by 0.8³ to get $7680
You get 0.8 because 100% minus 20% is 80% so you turn that into 0.8 and its 0.8³ because theres 3 years. Hope that made sense!
jeka943 years ago
8 0
\bf \qquad \textit{Amount for Exponential Decay}
\\\\
A=P(1 - r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &15000\\
r=rate\to 20\%\to \frac{20}{100}\to &0.20\\
t=\textit{elapsed time}\to &3\\
\end{cases}
\\\\\\
A=15000(1-0.20)^3\implies A=15000(0.80)^3
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What is my fav color/ what is statistical and non statistical
frez [133]

Answer:

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statistical is relating to the use of statistics.

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3 years ago
Help will give brainliest!!!!!!
sashaice [31]

Answer:

2.5

type A/B

Step-by-step explanation:

10/2.5=4

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3 years ago
PLEASE HELP ME. NO PHONY ANSWERS TY, ANSWER IF YOUR 100% RIGHT!
Ksivusya [100]

The relative frequency of female mathematics majors will be 0.5142.

<h3>How to find the relative frequency?</h3>

The proportion of the examined subgroup's value to the overall account is known as relative frequency.

A sample of 317 students at a university is surveyed.

The students are classified according to gender (“female” or “male”).

The table is given below.

Then the relative frequency of female mathematics majors will be

⇒ 36 / (36 + 34)

⇒ 36 / 70

⇒ 0.5142

Learn more about conditional relative frequency here:

brainly.com/question/8358304

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8 0
1 year ago
Price is $166.30 she has $7.08 how much more is needed
alexdok [17]
Subtract $7.08 from $166.30 = $159.22
8 0
3 years ago
Read 2 more answers
Which of the following graphs shows the solution set for the inequality below? 3|x + 1| &lt; 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.

3 0
2 years ago
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