Translate into an alegbraic expression and simplify if possible.

Na figura, r//s. O valor, em graus, de a+b é: a)180 b) 196 c) 216 d) 240

Nastasia [14]

216. hope this helps you.

4 0

3 years ago

One meteorite weighs 1.8 pounds. A second meteorite weighs 2.35 times as much as the first meteorite weighs. How much does the s

Grace [21]

The answer is 4.23 the second meteorite wieghts 4.23 pounds

8 0

3 years ago

4/5G + 42 = 2.8G + 18 Plz solve with an explanation

inysia [295]

Answer:

12

Step-by-step explanation:

Simplify \frac{4}{5}G

5

4

G to \frac{4G}{5}

5

4G

.

\frac{4G}{5}+42=2.8G+18

5

4G

+42=2.8G+18

Multiply both sides by 5.

4g + 210 = 14g + 90

Simplify

14G + 90 - 4G to 10G + 90

Subtract 90 from both sides.

210-90=10G210−90=10G

Simplify 210-90210−90 to 120120.

120=10G120=10G

Divide both sides by 1010.

\frac{120}{10}=G

10

120

=G

Simplify \frac{120}{10}

10

120

to 1212.

12=G12=G

Switch sides.

G=12G=12

SORRY ABOUT ALL THIS WORK...

3 0

2 years ago

Read 2 more answers

Water pours into a tank at the rate of 2000 cm3/min. The tank is cylindrical with radius 2 meters. How fast is the height of wat

Gennadij [26K]

Volume of water in the tank:

$V=\pi (2\,\mathrm m)^2h=\pi(200\,\mathrm{cm})^2h$

Differentiate both sides with respect to time t :

$\dfrac{\mathrm dV}{\mathrm dt}=\pi(200\,\mathrm{cm})^2\dfrac{\mathrm dh}{\mathrm dt}$

V changes at a rate of 2000 cc/min (cubic cm per minute); use this to solve for dh/dt :

$2000\dfrac{\mathrm{cm}^3}{\rm min}=\pi(40,000\,\mathrm{cm}^2)\dfrac{\mathrm dh}{\mathrm dt}$

$\dfrac{\mathrm dh}{\mathrm dt}=\dfrac{2000}{40,000\pi}\dfrac{\rm cm}{\rm min}=\dfrac1{20\pi}\dfrac{\rm cm}{\rm min}$

(The question asks how the height changes at the exact moment the height is 50 cm, but this info is a red herring because the rate of change is constant.)

7 0

3 years ago

Your friend asks if you would like to play a game of chance that uses a deck of cards and costs $1 to play. They say that if you

gtnhenbr [62]

Answer:

Expected value = 40/26 = 1.54 approximately

The player expects to win on average about $1.54 per game.

The positive expected value means it's a good idea to play the game.

============================================================

Further Explanation:

Let's label the three scenarios like so

scenario A: selecting a black card
scenario B: selecting a red card that is less than 5
scenario C: selecting anything that doesn't fit with the previous scenarios

The probability of scenario A happening is 1/2 because half the cards are black. Or you can notice that there are 26 black cards (13 spade + 13 club) out of 52 total, so 26/52 = 1/2. The net pay off for scenario A is 2-1 = 1 dollar because we have to account for the price to play the game.

-----------------

Now onto scenario B.

The cards that are less than five are: {A, 2, 3, 4}. I'm considering aces to be smaller than 2. There are 2 sets of these values to account for the two red suits (hearts and diamonds), meaning there are 4*2 = 8 such cards out of 52 total. Then note that 8/52 = 2/13. The probability of winning $10 is 2/13. Though the net pay off here is 10-1 = 9 dollars to account for the cost to play the game.

So far the fractions we found for scenarios A and B were: 1/2 and 2/13

Let's get each fraction to the same denominator

1/2 = 13/26
2/13 = 4/26

Then add them up

13/26 + 4/26 = 17/26

Next, subtract the value from 1

1 - (17/26) = 26/26 - 17/26 = 9/26

The fraction 9/26 represents the chances of getting anything other than scenario A or scenario B. The net pay off here is -1 to indicate you lose one dollar.

-----------------------------------

Here's a table to organize everything so far

$\begin{array}{|c|c|c|}\cline{1-3}\text{Scenario} & \text{Probability} & \text{Net Payoff}\\ \cline{1-3}\text{A} & 1/2 & 1\\ \cline{1-3}\text{B} & 2/13 & 9\\ \cline{1-3}\text{C} & 9/26 & -1\\ \cline{1-3}\end{array}$

What we do from here is multiply each probability with the corresponding net payoff. I'll write the results in the fourth column as shown below

$\begin{array}{|c|c|c|c|}\cline{1-4}\text{Scenario} & \text{Probability} & \text{Net Payoff} & \text{Probability * Payoff}\\ \cline{1-4}\text{A} & 1/2 & 1 & 1/2\\ \cline{1-4}\text{B} & 2/13 & 9 & 18/13\\ \cline{1-4}\text{C} & 9/26 & -1 & -9/26\\ \cline{1-4}\end{array}$

Then we add up the results of that fourth column to compute the expected value.

(1/2) + (18/13) + (-9/26)

13/26 + 36/26 - 9/26

(13+36-9)/26

40/26

1.538 approximately

This value rounds to 1.54

The expected value for the player is 1.54 which means they expect to win, on average, about $1.54 per game.

Therefore, this game is tilted in favor of the player and it's a good decision to play the game.

If the expected value was negative, then the player would lose money on average and the game wouldn't be a good idea to play (though the card dealer would be happy).

Having an expected value of 0 would indicate a mathematically fair game, as no side gains money nor do they lose money on average.

7 0

2 years ago