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AleksandrR [38]
2 years ago
9

I dont understand what to do and what the answer is

Mathematics
2 answers:
Mademuasel [1]2 years ago
4 0

Answer:

-1

Step-by-step explanation:

\frac{(4-2)}{(-2-0)}=\frac{2}{-2} \\\frac{2}{-2}= -1

Here is what I did to find the slope:

We will use the first two rows

The first row will be <em>x</em>₂ and <em>y</em>₂

The second row will be <em>x</em>₁ and <em>y</em>₁

The equation to find any slope is: \frac{y_{2}-y_{1} }{x_{2}-x_{1}}

hodyreva [135]2 years ago
3 0

Answer:

-1

Step-by-step explanation:

slope=(y2-y1)/(x2-x1)=(2-4)/(0+2)=-2/2=-1

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A rectangle has a width of 9 units and a length of 40 units. What is the length of a diagonal?
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Pythagoras tells us that the diagonal is √(9²+40²) = 41.
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3 years ago
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The area of a parallelogram is given by the formula A=bh. The area of the parallelogram is 85 sq. Units. Find it’s base and heig
kolbaska11 [484]

Answer:

the height is 5 and the base is 17.

Step-by-step explanation:

X + 1 is the height and 4x + 1 is the base, so we can substitute this into an equation:

( x + 1 ) (4x + 1) = 85

{apply the distributive law}

4x² + x + 4x + 1 =

4x² + 5x + 1 = 85

Take 85 to the other side by -85 each side:

4x² + 5x + 1 - 85 = 85 - 85

4x² + 5x - 84 = 0

{Now we have a quadratic equation}

The roots are either -5.25 or 4. X must be 4 since it must be positive as it is a side length.

Substitute x into the base and height to get the answers.

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Therefore the height is 5 and the base is 17.

7 0
2 years ago
Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of tim
Debora [2.8K]

Answer:

The Probability of exactly one tag being lost, in terms of π is 2\pi(1-\pi)-\pi^2(1-\pi)^2

Step-by-step explanation:

Using the tree diagram attached to the bottom of this answer, you can see that the probability of only one tag being lost is the union of the probability of the left tag being lost when the right one is not lost and the probability of the right tag being lost when the left one is not lost.

Probability of losing only the right tag:

P(C2|\frac{}{C1})=P(C2)*\frac{}{P(C1)}=\pi*(1-\pi)

Probabilty of losing only the left tag:

P(C1|\frac{}{C2})=P(C1)*\frac{}{P(C2)}=\pi*(1-\pi)

Now, to unite those two probabilities, we use basic probability properties:

P(C2|\frac{}{C1}) ∪ P(C1|\frac{}{C2})=P(C2|\frac{}{C1})+P(C1|\frac{}{C2})-(P(C1|\frac{}{C2})∩P(C2|\frac{}{C1}))

Since the events are independent:

P(C1|\frac{}{C2})∩P(C2|\frac{}{C1})=P(C1|\frac{}{C2})*P(C2|\frac{}{C1})

So, the union becomes:

P(C2|\frac{}{C1})∪P(C1|\frac{}{C2})=P(C1|\frac{}{C2})+P(C2|\frac{}{C1})-P(C1|\frac{}{C2})*P(C2|\frac{}{C1})

Replacing:

=\pi(1-\pi)+\pi(1-\pi)-\pi^2(1-\pi)^2=2\pi(1-\pi)-\pi^2(1-\pi)^2

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