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3 years ago

6

Desmond set his watch 16 seconds behind, and it falls behind another 1 second everyday. How far behind is Desmond's watch if he

last set it 14 days ago?

1 answer:

kondaur [170]3 years ago

3 0

Multiply how many seconds it falls behind a day by how many days have passed and you get your answer of 30 seconds

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What is 2/3 divided by 10

goldfiish [28.3K]

0.0666666667 I. ksbdbobfnfvrbdkbd

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3 years ago

The probability for event A is 0.3, the probability for event B is 0.6, and the probability of events A or B is 0.8. Why are the

Sedaia [141]

Answer:

A.

Step-by-step explanation:

I think it makes the most sense sorry if i got it wrong

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3 years ago

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Solve for x: 2x + 13 = 8x - 41

kherson [118]

Answer:

9 is your answer for x.

Step-by-step explanation:

Note the equal sign, what you do to one side, you do to the other. Isolate the variable, x. Do the opposite of PEMDAS.

PEMDAS is the order of operations, & =

Parenthesis

Exponent (& Roots)

Multiplication

Division

Addition

Subtraction

First, subtract 13 & 8x from both sides:

2x (-8x) + 13 (-13) = 8x (-8x) - 41 (-13)

2x - 8x = -41 - 13

Simplify. Combine like terms:

-6x = -54

Isolate the variable, x. Divide -6 from both sides:

(-6x)/-6 = (-54)/-6

x = (-54)/(-6)

x = 9

x = 9 is your answer.

~

8 0

3 years ago

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A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 y

Anestetic [448]

Answer:

$\frac{1}{4180}$

Step-by-step explanation:

GIVEN: A snack-size bag of M&Ms candies is opened. Inside, there are $12$ red candies, $12$ blue, $7$ green, $13$ brown, $3$ orange, and $10$ yellow. Three candies are pulled from the bag in succession, without replacement.

TO FIND: What is the probability that the first two candies drawn are orange and the third is green.

SOLUTION:

Total candies in the bag $=57$

Probability that first ball is orange, $P(A)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}$

$=\frac{3}{57}=\frac{1}{19}$

Probability that second ball is orange, $P(B)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}$

$=\frac{2}{56}=\frac{1}{28}$

Probability that third ball is green, $P(C)=\frac{\text{total green candies in bag}}{\text{total candies in bag}}$

$=\frac{7}{55}$

Now, probability that first two balls are orange and third is green is

$=P(A)\times P(B)\times P(C)$

$=\frac{1}{19}\times\frac{1}{28}\times\frac{7}{55}$

$=\frac{1}{19}\times\frac{1}{4}\times\frac{1}{55}$

$=\frac{1}{4180}$

Hence, probability that first two balls are orange and third is green is $\frac{1}{4180}$

3 0

4 years ago