A dice is rolled, then a coin is tossed. What is the probability of getting a 5 then tails? Only fractions, NO decimals. *

Mathematics

2 answers:

Rainbow [258]2 years ago

7 0

Answer:

1/12

Step-by-step explanation:

The probability of getting a 5 is 1/6. The probability of getting tails is 1/2. Multiply 1/6 by 1/2 and you will get 1/12

katrin2010 [14]2 years ago

6 0

Answer:

1/12

The probability of getting a 5 is 1/6. The probability of getting tails is 1/2. Multiply 1/6 by 1/2 and you will get 1/12

Step-by-step explanation:

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Sam and Liam raced each other up and then down the hill. Sam's average speed up the hill was 1 mph, and his average speed down w

Furkat [3]

Answer:

Sam's time was 1 hour and 6.66 minutes (or 1 and 1/9 hours)

Liam's time was was 1 hour

So in conclusion, Liam was faster

Step-by-step explanation:

We got Sam's time by 1/1 because Sam was walking 1 mph and the length to reach the top of the hill is 1 mile. So it would take Sam 1 hour to walk 1 mile. Walking down the hill Sam's speed was 9 mph. So the equation would be 1/9. So you would get 1/9. 1/9 of 60 is 6.66 repeating.

For Liam it is pretty simple. Since he is going up the hill at 2 mph and the length of the bottom of the hill to the top is 1 mile, it would be 30 minutes. Repeat this for the top to the bottom and in total it would be 1 hour.

Hope this helps

3 0

2 years ago

A closed box has a square base with side length feet and height feet given that the volume of the box is 34 ft.³ express the sur

Deffense [45]

Answer:

$\dfrac{2s^3+136}{s}$

Step-by-step explanation:

Let the side length of the square base =s feet

Let the height of the box = h

Given that the volume of the box = $34$ ft^3$

Volume of the box = $s^2h$

Then:

$s^2h=34$ ft^3\\$Divide both sides by s^2\\h=\dfrac{34}{s^2}$

Surface Area of a Rectangular Prism =2(lb+bh+lh)

Since we have a square base, l=b=s feet

Therefore:

Surface Area of our closed box $= 2(s^2+sh+sh)$

$S$urface Area= 2s^2+4sh\\Recall: h=\dfrac{34}{s^2}\\$Surface Area= 2s^2+4s\left(\dfrac{34}{s^2}\right)\\=2s^2+\dfrac{136}{s}\\$Surface Area in terms of length only=\dfrac{2s^3+136}{s}$