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3 years ago

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checking account a charges a montly service fee of 3 and a per check fee of .02 while checking account b charges a montly servic

e fee of 2 and a per check fee of .03 how many checks would a person have to write for the two accounts to cost the same

1 answer:

Gnoma [55]3 years ago

3 0

Answer:

good morning I was just asking when will I be

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An angle bisector of a triangle divides the opposite side of the triangle into segments 1.5 cm and 1 cm long. A second side of t

bagirrra123 [75]

By the angle bisector theorem, one of the lengths of the third side of the triangle is given by:
1.5 / 1 = x / 2.6
x = 1.5 x 2.6 = 3.9
One of the possible length of the third side is 3.9 cm

The other possible length is given by 1 / 1.5 = x / 2.6
1.5x = 2.6
x = 2.6/1.5 = 1.7
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2 years ago

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.

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