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2 years ago

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At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 31% of courses have a res

earch paper and a final exam. Find the probability that a course has a final exam or a research paper.

1 answer:

beks73 [17]2 years ago

6 0

Answer:

0.25

Step-by-step explanation:

72% of courses have final exams and 46% of courses require research papers which means probability of 0.72 for courses that have final exams and 0.46 for courses that require research papers.

31% of courses have a research paper and a final exam, which means probability of 0.31 for both courses with exams and research papers, using Venn diagram approach, find picture attached to the solution.

P(R or E) = P(R) + P(E) - P(R and E), which gives:

P(R or E) = 0.15 + 0.41 - 0.31

P(R or E) = 0.25.

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Which answer is the same as moving the decimal in 36.2 four places to the left?

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Answer:

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Step-by-step explanation:

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2 years ago

A point x is 34m due to east of a point Y . The bearing of a flagpole from x and Y are N18°W And N40°E respectively .

Murljashka [212]

Answer:

The distance of flag post from Y is 38.13 m

Step-by-step explanation:

Consider point Y at the intersection of both lines as shown below. Now the point X lies 34 meter from point Y in east direction.

Now flag pole at point X lies at a bearing of N18°W. That is at point X from north, flag post makes an angle of 18° towards west.

Similarly flag pole at point Y lies at a bearing of N40°E. That is at point Y from north, flag post makes an angle of 40° towards east.

Consider ∆ AXY as right angle triangle. Therefore measure of angle FXY is,

$\angle AXY=\angle AXF+\angle FXY$

$\therefore 90 \degree=18\degree+\angle FXY$

$\angle FXY=72\degree$

Consider ∆ BYX as right angle triangle. Therefore measure of angle FYX is,

$\angle BYX=\angle BYF+\angle FYX$

$\therefore 90 \degree=40\degree+\angle FYX$

$\angle FYX=50\degree$

Refer attachment 1.

From diagram consider the triangle FYX. To find the third angle that is ∠YFX can be calculated by using angle sum property of triangle.

∠YFX+∠FYX+∠FXY=180°

∠YFX+50°+72°=180°

∠YFX=58°

Refer attachment 2.

Now the distance FY can be calculated using sin rule as follows,

$\frac{\sin X}{FY}=\frac{\sin F}{YX}=\frac{\sin Y}{FX}$

Substituting the values,

$\frac{\sin 72}{FY}=\frac{\sin 58}{34}=\frac{\sin 50}{FX}$

Simplifying first two terms,

$\frac{\sin 72}{FY}=\frac{\sin 58}{34}$

Cross multiplying,

$34\times\sin 72=FY\times \sin 58$

$\dfrac{34\times\sin 72}{\sin 58}=FY$

$FY=38.13$ m

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2 years ago

The list shows the area in square feet of each apartment available for rent in a building. 565, 961, 867, 517, 627, 714, 517, 72

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STEP 1:
put numbers in order from lowest to highest
517, 517, 565, 627, 714, 728, 867, 961

STEP 2:
find the difference between the lowest and highest numbers

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ANSWER: The range of the areas is 444 ft^2.

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2 years ago

Read 2 more answers

Two runners are saving money to attend a marathon. The first runner has $112 in savings, received a $45 gift from a friend, and

scoundrel [369]

Answer:

The correct answer is the last option, that is, $112 + 25m + 45 = 50 + 60m$

Step-by-step explanation:

We have been given that the first runner has $112 in savings, received a $45 gift from a friend, and will save $25 each month. Therefore, amount of money in the account of first running after m months will be: $112+45+25m$

We have been given that the second runner has $50 in savings and will save $60 each month. Therefore, amount of money in the account of second running after m months will be: $50 + 60m$

In order for amount of money to be equal in accounts of both the runners, we set up:

$112+45+25m=50+60m$

Upon rewriting the left hand side using commutative law, we get:

$112+25m+45=50+60m$

Therefore, we can see that the last option is the correct answer.

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2 years ago

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A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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1 year ago