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3 years ago

PLEASE HELP ME OUT HERE!!!!

Mathematics

1 answer:

aleksandr82 [10.1K]3 years ago

6 0

It should be the second option i’m positive

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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

3 years ago

PLZZZZZZZZZZZZZZZ HELPPPPP!!! I"LLL BUY YOU CHOCOLATE!!!

shtirl [24]

Answer:

The answer is 83

I need CHOCOLATE.

8 0

3 years ago

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40 Points correct answer

KIM [24]

<span>8(-y+5)+6(2y-7)
= -8y + 40 + 12y - 42 ....expand by using distributive property
= 4y - 2 ...............combine like terms

hope it helps</span>

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3 years ago

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A family built a house in a beach resort area. In 1990, the house was 430 ft from the water. With erosion, the house was 400 ft

neonofarm [45]

We develop an equation for the given situation by first writing the general equation for lines,

y = mx + b

Substituting to this given the values given above,

(1990) 430 = b

(2000) 400 = m(10) + 430

The value of m from the equation in 2000 is -3. Thus, the equation of that relates the variables is,

y = -3x + 430