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4 years ago

I think of a number multiply by 3,add 4,i get 22

Mathematics

2 answers:

MrRissso [65]4 years ago

7 0

Answer:

Step-by-step explanation:

22 - 4 = 18

18 / 3 = 6

Katyanochek1 [597]4 years ago

3 0

Answer:

Step-by-step explanation:

22-4=18/3=6

start from what you get that use the inverse to get the the original answer

You might be interested in

Indicate the equation of the line that is the perpendicular bisector of the segment with endpoints (4, 1) and (2, -5).

PtichkaEL [24]

Okay to find the perpendicular bisector of a segment you first need to find the slope of the reference segment.

m=(y2-y1)/(x2-x1) in this case:

m=(-5-1)/(2-4)

m=-6/-2

m=3

Now for the the bisector line to be perpendicular its slope must be the negative reciprocal of the reference segment, mathematically:

m1*m2=-1 in this case:

3m=-1

m=-1/3

So now we know that the slope is -1/3 we need to find the midpoint of the line segment that we are bisecting. The midpoint is simply the average of the coordinates of the endpoints, mathematically:

mp=((x1+x2)/2, (y1+y2)/2), in this case:

mp=((4+2)/2, (1-5)/2)

mp=(6/2, -4/2)

mp=(3,-2)

So our bisector must pass through the midpoint, or (3,-2) and have a slope of -1/3 so we can say:

y=mx+b, where m=slope and b=y-intercept, and given what we know:

-2=(-1/3)3+b

-2=-3/3+b

-2=-1+b

-1=b

So now we have the complete equation of the perpendicular bisector...

y=-x/3-1 or more neatly in my opinion :P

y=(-x-3)/3

4 0

3 years ago

Read 2 more answers

Suppose box A contains 4 red and 5 blue poker chips and box B contains 6 red and 3 blue poker chips. Then a poker chip is chosen

sergejj [24]

Answer:

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Coin chosen from box B is red.

Event B: Blue poker chip transferred.

Probability of choosing a red coin:

7/10 of 4/9(red coin from box A)

6/10 of 5/9(blue coin from box A). So

$P(A) = \frac{7}{10}*\frac{4}{9} + \frac{6}{10}*\frac{5}{9} = \frac{28 + 30}{90} = 0.6444$

Blue chip transferred, red coin chosen:

6/10 of 5/9. So

$P(A \cap B) = \frac{6}{10}*\frac{5}{9} = \frac{30}{90} = 0.3333$

What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3333}{0.6444} = 0.5172$

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

5 0

3 years ago

What is the interquartile range of the following data set? 3,8, 14, 19, 22, 29, 33, 37, 43, 49. A. 26 B. 23 C. 19 D. 18

Fynjy0 [20]

Answer:

I think the answer is B.23

8 0

3 years ago

OK. these questions are worth ALL my points (55) !

Ratling [72]

Answer:

1. -7

2. 212

3. -49

4.-1,991

5. 78

6. -16

7. 2

8. 100

9. 16

10. 16

11. 13

12. 75

13. |12| = 12

14.|12| = 12

15. |8|

16.| 10|

17.15

18. 1,400

Step-by-step explanation:

6 0

3 years ago

There are 7 red candies in a pack, 9 blue candies in a pack, and 12 pink candies in a pack. If your wonderful math teacher wants

pentagon [3]

Given:
7 red candies
9 blue candies
12 pink candies

Condition:
Number of bags that has equal number of candies.

We can do 7 bags with 1 colored candy each. There will be an excess of 2 blue candies and 5 pink candies.

We can do 4 bags with 1 red candy, 2 blue candies, 3 pink candies. There will be an excess of 3 red candies and 1 blue candy.

We can do 3 bags with 2 red candies, 3 blue candies, 4 pink candies. There will be an excess of 1 red candy.

5 0

3 years ago