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3 years ago

I have 7 candles lit 2 go out how many do i have left

1 answer:

8 0

If you have 7 candles lit and 2 go out, the number of candles left is given by

7 - 2 = 5

Therefore, 5 candles are left lit.

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Anyone know the answer to this algebra question?

lana [24]

Answer:

possibly the third maybe

4 0

3 years ago

I need help on questions two lol pls help me

aksik [14]

1. (8/9)/4

8/9 * 1/4 = 8/36= 2/9 paint for each birdhouse

2. ron 1 mile every 20 minutes so 3 mph

1 mile every 24 minutes so ron is the faster walker

5 0

2 years ago

What is the solution to -2(8x – 4) < 2x + 5? він x x оооо x x заще х: 6 о

lozanna [386]

Answer:

The answer to the question is x<3/18

7 0

3 years ago

I roll a fair die twice and obtain two numbers X1= result of the first roll and X2= result of the second roll. Given that I know

azamat

By definition of conditional probability,

$P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{P((X_1=4\text{ or }X_2=4)\text{ and }X_1+X_2=7)}{P(X_1+X_2=7)}$

$=\dfrac{P((X_1=4\text{ and }X_1+X_2=7)\text{ or }(X_2=4\text{ and }X_1+X_2=7))}{P(X_1+X_2=7)}$

Assuming a standard 6-sided fair die,

if $X_1=4$ , then $X_1+X_2=7$ means $X_2=3$ ; otherwise,
if $X_2=4$ , then $X_1=3$ .

Both outcomes are mutually exclusive with probability $\frac1{36}$ each, hence total probability $\frac2{36}=\frac1{18}$ .

Of the 36 possible outcomes, there are 6 ways to sum the integers 1-6 to get 7:

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

and so a sum of 7 occurs $\frac6{36}=\frac16$ of the time.

Then the probability we want is

$P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{\frac1{18}}{\frac16}=\frac13$

6 0

3 years ago

Mike has a bag full of M&M's. In the bag, there are three green M&M's, two blue M&M's, one red M&M, one brown M&

astraxan [27]

3/8

6/7
Hope this helps!

7 0

3 years ago