Ask question

Ask question

All categories

3 years ago

8

Jim has 6 quarters, 1 dime, 5 nickels, and 2 pennies in his pocket. If one coin falls from his pocket, what is the probability t

hat the coin is a quarter or a penny?

1 answer:

alexdok [17]3 years ago

8 0

The probability that a quarter will fall is 6/14 or 3/7 simplified. The probability that a penny will fall is 2/14 or 1/7 simplified.

You might be interested in

Bailey spent $12.25 at the arcade on Saturday and $9.50 at the arcade on Sunday. How much money did Bailey spend anogether?

galben [10]

Answer:

21.75

Step-by-step explanation:

i added 12.25 by 9.50

8 0

3 years ago

Which temperature is colder? -10 degrees or -13 degrees

frez [133]

Answer:

-13

Step-by-step explanation:

The more negative the number, the colder it is

-13 is more negative than -10

-13 < -10

-13 is colder

3 0

2 years ago

What is m∠B?<br><br>A.51<br><br>B..129<br><br>C.134<br><br>D.141

san4es73 [151]

Answer:

I think it 129 If not try 4

7 0

3 years ago

How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?

MrRissso [65]

$\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3} \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(3x)^{3 \times \frac{5}{3} }}{(2y^3)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}$

Now, If the exponent was negative like you asked....

$\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}$

5 0

3 years ago

PLEASE HELP! Thank you!

Lilit [14]

T⁻¹ is the inverse function of T.
To get the inverse function, replace x with y and y with x.
So T<span>⁻¹(x, y): (x + 2, y - 7)</span>

7 0

3 years ago