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3 years ago

To start a board game,the player who rolls two sixes gets to go first. What is the probability that a player will roll two

1 answer:

5 0

There is a 1/36 or a ~2.8% chance that a player will roll two sixes. Because the dice is cubed, the player will have a 1/6 probability of rolling a six one time. Because the player needs to get six 2 times in a row, you would simply need to multiply 1/6 by itself twice(1/6•1/6 or 1/6^2) which equals 1/36.

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I need help<br> <img src="https://tex.z-dn.net/?f=%5Cbf%203x-5%3D16" id="TexFormula1" title="\bf 3x-5=16" alt="\bf 3x-5=16" alig

photoshop1234 [79]

Answer: x = 7

Step-by-step explanation:

do opposites to both sides. what is the opposite of -5? it would be plus 5.

add 5 to both sides

3x = 21

now what is the opposite of 3 times x? 3 divided by.

divide 3 to both sides

x = 7

4 0

3 years ago

Read 2 more answers

Solve the system of equations and choose the correct ordered pair.

konstantin123 [22]

The answer is A (9,4)

5 0

3 years ago

Read 2 more answers

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

4 years ago

What is the interquartile range of the following set of numbers? 114, 90, 83, 101, 97, 142, 117, 87, 72

Likurg_2 [28]

Hello!

First we have to find the median

We have to order the numbers from least to greatest

72, 83, 87, 90, 97, 101, 114, 117, 142

The median is 97

Next we have to find quartile 1

you look for the median for the numbers under 97

It is 83 and 87

You get the average

83 + 87 = 170

170 / 2 = 85

quartile 1 is 85

Next we find quartile 3

Look for the median for the numbers above 97

The numbers are 114 and 117

Get the average

114 + 117 = 231

231/2 = 115.5

Now you find the range between the quartiles

You do quartile 3 - quartile 1

115.5 - 85 = 30.5

The answer is 30.5

Hope this helps!

5 0

4 years ago

Read 2 more answers

HELP ASAP 15 POINTS

Elena-2011 [213]

It would be 1/3 then A,D,C,5/3,A respectively

4 0

3 years ago