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Masteriza [31]
3 years ago
10

Hey can you please help me posted picture of question

Mathematics
1 answer:
Oksanka [162]3 years ago
5 0
The statement is true.

P(A|B) is the probability of occurrence of event A, provided that(given that) event B has already occurred.

This is known as conditional probability. In conditional probability, the event on right side of the vertical bar (which is B in this case) is given to have already occurred (either we assume this, or some evidence is given about this) and we calculate the probability of event on left of the vertical bar (which is A in this case) based on this information. The formula of condition probability is:

P(A|B)= \frac{P(A*B)}{P(B)} 

P(A*B) indicates the probability of intersection of event A and B.

So the correct answer is TRUE.
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T -10 -9 -8-7-6-5-4-3-2-1. 0 1 2 3 4 5 6 7 8 9
Ainat [17]
The correct answer is C !
7 0
3 years ago
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HUDSON HAS TO FILL A POT WITH 1 QUART OF WATER. ALL HE HAS IS 1/4 CUP MEASURING CUP. HOW MANY TIMES DOES HE HAVE TO FILL THE 1/4
Marta_Voda [28]
There are 4 cups in a quart.

So if Hudson has only a 1/4 measuring cup, this can be represented by the equation:

4/1/4, solving it we get: 4* 4 (dividing turns a fraction into its reciprocal)

So, 4*4=16

Hudson will have to fill the 1/4 measuring cup 16 times to get a quart.

I hope this helps!
4 0
3 years ago
Read 2 more answers
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.

3 0
2 years ago
No step by step answers or links
inysia [295]

Answer:

1. k = 15

2. d = 20

3. n = 3

6 0
3 years ago
Please help me! Im making the first correct answer brainliest!
masha68 [24]

Answer:

You didn't ask a question.

Step-by-step explanation:

7 0
3 years ago
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