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2 years ago

Nancy has 49 pennies in her bank

Mathematics

1 answer:

lubasha [3.4K]2 years ago

5 0

what’s the rest of the question??

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Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco

kari74 [83]

Answer:

C. $\frac{1}{18}$

Step-by-step explanation:

Given: Six cards numbered from $1$ to $6$ are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is $8$ , what is the probability that one of the two cards drawn is numbered $5$ .

Solution:

Sample space for sum of cards when two cards are drawn at random is $\{(1,1),(1,2),(1,3)......(6,6)\}$

total number of possible cases $=36$

Sample space when sum of cards is $8$ is $\{(3,5),(5,3),(6,2),(2,6),(4,4)\}$

Total number of possible cases $=5$

Sample space when one of the cards is $5$ is $\{(5,3),(3,5)\}$

Total number of possible cases $=2$

Let A be the event that sum of cards is $8$

$p(\text{A})$ $=\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}$

$p(\text{A})=\frac{5}{36}$

Let B be the event when one of the two cards is $5$

probability than one of two cards is $5$ when sum of cards is $8$

$p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}$

$p(\frac{\text{B}}{\text{A}})=\frac{2}{5}$

Now,

probability that sum of cards $8$ is and one of cards is $5$

$p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})$

$p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}$

$p(\text{A and B})=\frac{1}{18}$

if sum of cards is $8$ then probability that one of the cards is $8$ is $\frac{1}{18}$ , option C is correct.

3 0

2 years ago

Please help. It's algebra

spayn [35]

Answer: For the first option it is 28y^2 +31y - 10

Step-by-step explanation: Hope this helps

7 0

2 years ago

Find the missing term and the missing coefficient 6a −( )5a =( ) a2 − 35a

andrew11 [14]

Solving for the missing term and the missing coefficient (6a − )5a = ( ) a^2 − 35a
Let the missing term be X
Let the missing coefficient be Y
Therefore, (6a – X)5a = Y(a^2) – 35a
6a x 5a – X.5a = Y.a^2 – 35a
30a^2 – X.5a = Y.a^2 – 35a
Equating co-efficients,
30a^2 = Y.a^2; X.5a = 35a
30 = Y; 5X = 35
Y = 30; X = 7
Therefore, (6a-7)5a = 30 a^2 – 35a <span>
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3 0

2 years ago

Which is the logarithmic form of 9^2 = 81?

pochemuha