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3 years ago

6

How many soulutions does -2(6-3x)=-12+6x have ?

1 answer:

Natalija [7]3 years ago

5 0

Answer:

Step-by-step explanation:

Only one

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Two random variables x and y are independent if the value of x does not affect the value of y. If the variables are not indepen

jeyben [28]

Answer: $\mu_{x+y} =$ 3026

$\mu_{x-y}=$ 30

Step-by-step explanation: Average sum of the female and male's test score is the sum of expected value of each gender:

$\mu_{x+y}=\mu_{x}+\mu_{y}$

Assuming x represents the random male selected and y represents the random female selected:

$\mu_{x+y}=1528+1498$

$\mu_{x+y}=$ 3026

The average sum of their scores is 3026.

Average difference is the difference between the expected value (mean) of each gender:

$\mu_{x-y}=\mu_{x}-\mu_{y}$

$\mu_{x-y}=$ 1528 - 1498

$\mu_{x-y}=$ 30

The average difference of their scores is 30.

6 0

3 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.

3 0

3 years ago

Vertex A in quadrilateral ABCD lies at (-3, 2). If you rotate ABCD 180° clockwise about the origin, what will be the coordinates

natali 33 [55]

The rule for a rotation by 180° about the origin is (x,y) -->(−x,−y)

So A(-3, 2) to A'(3, -2)

Answer:

A'(3, -2)

4 0

3 years ago

PLEASE HURRY!! TIMED!! WILL GIVE BRAINLIEST!!!

tensa zangetsu [6.8K]

Answer:

2/3 +9.26 = 0.6666+9.26=rational

4 0

3 years ago

PLEASE HELP WILL MARK BRAINLIEST!

pantera1 [17]

Answer:

Make a coordinate plane and you have to solve around the origin say 4,4 and 4,1.

Step-by-step explanation:

8 0

3 years ago