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3 years ago

What is the probability that 2 randomly selected months have 31 days?

Mathematics

2 answers:

aniked [119]3 years ago

6 0

Answer:

31.8%

Step-by-step explanation:There are 7 months that have 31 days so,

The odds the first selected month has 31 days is 7/12

The next month there are only 6 months left with 31 days of only 11 months to choose from

so the odds for the second are 6/11

To calculate the odds they BOTH have 31 months you multiply the two odds:

(7/12) x (6/11) = 42/132 = .318 so 31.8%

belka [17]3 years ago

5 0

Answer:

6/11

Step-by-step explanation:

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<h3><u>The x intercept is at (-7, 0).</u></h3><h3><u>The y intercept is at (0, 1).</u></h3>

To find both the x and the y intercept, we need to solve one at a time.

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Y= x^2 -5 solve for x

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Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

<u>Transverse axis</u>

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

<u>The vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

<u>The co-vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola