<em>Answer</em><em>:</em><em> </em><em>3</em><em>7</em>
<em>Step</em><em> </em><em>by</em><em> </em><em>step</em><em> </em><em>explanation</em><em>:</em>
<em>y</em><em>+</em><em>2</em><em>9</em><em>+</em><em>4</em><em>0</em><em>+</em><em>2</em><em>y</em><em>=</em><em>1</em><em>8</em><em>0</em><em>°</em><em>(</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>angle</em><em> </em><em>in</em><em> </em><em>stra</em><em>ight</em><em> </em><em>line</em><em>)</em>
<em>or</em><em>,</em><em> </em><em>y</em><em>+</em><em>2</em><em>y</em><em>+</em><em>2</em><em>9</em><em>+</em><em>4</em><em>0</em><em>=</em><em>1</em><em>8</em><em>0</em><em>°</em>
<em>or</em><em>,</em><em>3</em><em>y</em><em>+</em><em>6</em><em>9</em><em>=</em><em>1</em><em>8</em><em>0</em>
<em>or</em><em>,</em><em>3</em><em>y</em><em>=</em><em>1</em><em>8</em><em>0</em><em>-</em><em>6</em><em>9</em>
<em>or</em><em>,</em><em>3</em><em>y</em><em>=</em><em>1</em><em>1</em><em>1</em>
<em>or</em><em>,</em><em>y</em><em>=</em><em>1</em><em>1</em><em>1</em><em>/</em><em>3</em>
<em>y</em><em>=</em><em>3</em><em>7</em>
<em>hope</em><em> </em><em>it</em><em> </em><em>helps</em>
<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>
Answer:
Kindly check explanation
Step-by-step explanation:
Total Number of pens = 5
Number of defective pens = 2
Number of non-defective pens = 3
A.) number of defective pens selected :
X : {0, 1, 2}
It is possible that no defective pen will be selected ; 1 defective will be chosen or both pens are defective.
2.)
Defective as Success (since selecting a defective pen is the point of interest.
3.)
Since selection is done without replacement
Probability of success per selection is different for each selection ;
Number of defective = 2
Number of observations = 5
P(success on first selection) = 2/5
P(success on second selection) = 1/4
Hence, X is not well approximated by a binomial random variable.
4.) if selection is done with replacement ; then then the probability of success per selection will be the Same for each selection made. Hence, X will be well approximated by a binomial Random variable.
5.) If sampling is done without replacement, then the the hypergeometric function will be a more effective approximation.
A parabola with an equation, y2 = 4ax has its vertex at the origin and opens to the right.
It's not just the '4' that is important, it's '4a' that matters.
This type of parabola has a directrix at x = -a, and a focus at (a, 0). By writing the equation as it is, the position of the directrix and focus are readily identifiable.
For example, y2 = 2.4x doesn't say a great deal. Re-writing the equation of the parabola as y2 = 4*(0.6)x tells us immediately that the directrix is at x = -0.6 and the focus is at (0.6, 0)
If you think about it, a hexagon can be cut into 6 equilateral triangles. Because of this, all you need to do is find the area of one triangle and multiply it by 6. Since the side length of one triangle is 12, we know that is the base. Since the triangles that are created when cutting the equilateral triangle in half are both 30 60 90, the height must be 6(3)^1/3. We can then multiply 12 • 6 radical 3 •1/2 to get 36 radical 3. Multiply this by 6 and get 216 radical 3 units squared.