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1 year ago

If two 6-sided dice are rolled what is the probability that both numbers will be even?

1 answer:

3 0

The probability of one dice rolling an even number is 1/2 as half the numbers are even. Since the events are independent, and there are 2 dice, we need to square 1/2 to get 1/4, assuming they are both fair dice

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At a bargain store, Tanya bought 4 items that each cost the same amount. Tony bought 5 items that each cost the same amount, but

Alex_Xolod [135]

If x is the amount that Tanya paid for each item, then Tony's items would be x-2.25. So:

3x=4(x-2.25)

3x=4x-9

x=9

x-2.25=6.75

Tanya spent $9 on each item; Tony spent $6.75 on each of his items. ☺☺☺☺

6 0

3 years ago

Help with 4 Multiple-choice

kykrilka [37]

Answer:

1. D

2. A

3. C

4. A

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

I WILL MARK BRAINLIEST

vekshin1

Hello!

You can use the Pythagorean Theorem

$a^{2} + b^{2} = c^{2}$

c is the hypotenuse which is the side across the right angle
a and b are the other sides

Put in the values you know

$a^{2} + 5^{2} = 12^{2}$

Square the numbers

$a^{2} + 25 = 144$

Subtract 25 from both sides

$a^{2} = 119$

Take the square root of both sides

a = 10.9

The answer is D) 10.9 centimeters

Hope this helps!

3 0

2 years ago

Read 2 more answers

You go shopping at a local department store with your friend and cousin. You buy one pair of jeans, four pairs of shorts, and tw

VashaNatasha [74]

Answer: Cost per pair of jeans = $ 20

Cost per pair of shorts = $12

Cost per shirt = $8

Step-by-step explanation:

Let x= cost per pair of jeans

y = Cost per pair of shorts

z= cost per shirt.

For you, total cost = x+4y+2z= 84

For your friend, total cost = 2x+y+3z=76

For cousin, total cost = x+2y+z= 52

Required system:

$x+4y+2z=84\ (i)\\ 2x+y+3z=76\ (ii)\\ x+2y+z=52\ (iii)$

from (i), $x=84-4y-2z$ (*)

put this in (ii) and (iii) we get

$2\left(84-4y-2z\right)+y+3z=76\\\Rightarrow\ -7y-z+168=76\ (iv)\\\\ 84-4y-2z+2y+z=52\\\Rightarrow\ -2y-z+84=52 \ (v)$

from (iv), $y=-\frac{z-92}{7}\ (vi)$

$\mathrm{Subsititute\:}y=-\frac{z-92}{7}\ \text{in}\ (v)$

$-2\left(-\frac{z-92}{7}\right)-z+84=52\\\\\Rightarrow\ -2\left(-\frac{z-92}{7}\right)-z=-32\\\\\Rightarrow\ -\frac{5z}{7}-\frac{184}{7}=-32\\\\\Rightarrow\ -5z-184=-224\\\\\Rightarrow\ -5z=-40\\\\\Rightarrow\ z=8$