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Kryger [21]
3 years ago
9

A cake recipe suggests a cooking temperature of 325 degree Fahrenheit if the batter is less than one inch thick and 350F if the

batter is more than one inch thick. If the cake batter has a volume of 100 in^3 and your cake pan is 9inches by 11 inches, what does the oven need to set at if the oven temperature settings are in degrees Celsius? (round to nearest whole number)
Mathematics
1 answer:
bija089 [108]3 years ago
6 0
The first step is finding the height of the cake. The formula for the volume of a box: Volume= length x width x height. So given the volume and length and width: 100= 9 x 11 x h. 100=99h and solve for h. h= 1.01 inches so the batter is more than one inch thick. That means that the matter needs to be baked at 350 degrees Fahrenheit. The formula to convert from Fahrenheit to Celsius:  T(C)= (T(F)-32) x (5/9). So (350-32) x (5/9) = 176.667. Rounded to the nearest whole number is 177 degrees celsius. 
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a) No, because you have only 33.8% of chances of winning the bet.

b) No, because you have only 44.7% of chances of winning the bet.

Step-by-step explanation:

a) Of the total amount of cards (n=52 cards) there are 12 face cards (3 face cards: Jack, Queen, or King for everyone of the 4 suits: clubs, diamonds, hearts and spades).

The probabiility of losing this bet is the sum of:

- The probability of having a face card in the first turn

- The probability of having a face card in the second turn, having a non-face card in the first turn.

- The probability of having a face card in the third turn, having a non-face card in the previous turns.

- The probability of having a face card in the fourth turn, having a non-face card in the previous turns.

<u><em>1) The probability of having a face card in the first turn</em></u>

In this case, the chances are 12 in 52:

P_1=P(face\, card)=12/52=0.231

<u><em>2) The probability of having a face card in the second turn, having a non-face card in the first turn.</em></u>

In this case, first we have to get a non-face card (there are 40 in the dech of 52), and then, with the rest of the cards (there are 51 left now), getting a face card:

P_2=P(non\,face\,card)*P(face\,card)=(40/52)*(12/51)=0.769*0.235=0.181

<u><em>3) The probability of having a face card in the third turn, having a non-face card in the first and second turn.</em></u>

In this case, first we have to get two consecutive non-face card, and then, with the rest of the cards, getting a face card:

P_3=(40/52)*(39/51)*(12/50)\\\\P_3=0.769*0.765*0.240=0.141

<u><em>4) The probability of having a face card in the fourth turn, having a non-face card in the previous turns.</em></u>

In this case, first we have to get three consecutive non-face card, and then, with the rest of the cards, getting a face card:

P_4=(40/52)*(39/51)*(38/50)*(12/49)\\\\P_4=0.769*0.765*0.76*0.245=0.109

With these four probabilities we can calculate the probability of losing this bet:

P=P_1+P_2+P_3+P_4=0.231+0.181+0.141+0.109=0.662

The probability of losing is 66.2%, which is the same as saying you have (1-0.662)=0.338 or 33.8% of winning chances. Losing is more probable than winning, so you should not take the bet.

b) If the bet involves 3 cards, the only difference with a) is that there is no probability of getting the face card in the fourth turn.

We can calculate the probability of losing as the sum of the first probabilities already calculated:

P=P_1+P_2+P_3=0.231+0.181+0.141=0.553

There is 55.3% of losing (or 44.7% of winning), so it is still not convenient to bet.

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