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3 years ago

Assume there are 365 days in a year.

1 answer:

4 0

The chance of student 1's birthday being individual is 365/365 or 100%.
Then the chance of student 2's birthday being different is 364/365.
Then it's narrowed down to 363/365 for student 3 and so on until you get all 10 students.
If you multiply all these values together, the probability would come out at around 0.88305182223 or 0.88.

To get all the same birthday you'd have to the chance of one birthday, 1/365 and multiply this by itself 10 times. This will produce a very tiny number. In standard form this would be 2.3827x10'-26 or in normal terms: 0.23827109210000000000000000, so very small.

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Can somebody plz answer these questions correctly thanks!!

nadya68 [22]

Answer:

The decimal form of 3/5 is .6 or .60 and the percent form is 60%.

The fraction form of 0.3 is 30/100 or 3/10 and the percent form is 30%.

The decimal form of 3/20 is .15 and the percent form is 15%.

The fraction form of 0.58 is 58/100 or 29/50 and the percent form is 58%.

The decimal form of 19/50 is .38 and the percent form is 38%

4 0

3 years ago

Distribute property to simplify 4(3x+2)

Julli [10]

Here is the equation:

$4(3x+2)$

Since 4 is next to the parentheses, you have to multiply everything in the parentheses by 4.

First multiply 3x by 4:

$3x \times 4 = 12x$

Now multiply 2 by 4:

$4 \times 2 = 8$

Now add them together:

$12x+8 = 12x +8$

Your answer is <u>12x + 8</u>

8 0

3 years ago

How good are you setting goals? 3 sentences for honors

Lunna [17]

Answer:

Be more productive

Step-by-step explanation:

This could be a goal

5 0

3 years ago

The school owns 2 classroom sets of 30 calculators each, which students are required to have during their mathematics class. The

sattari [20]

There's not enough information here.

7 0

3 years ago

Read 2 more answers

How would you draw a counterexample to this?

NeX [460]

Melanie said:
Every angle bisector in a triangle bisects the opposite side perpendicularly.

A 'counterexample' would show an angle bisector in a triangle that DOESN'T
bisect the opposite side perpendicularly.

See my attached drawing of a counterexample.

Both of the triangles that Melanie examined have equal sides on both sides
of the angle bisector. That's the only way that the angle bisector can bisect
the opposite side perpendicularly. Melanie didn't examine enough different
triangles.

6 0

3 years ago