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Scrat [10]
3 years ago
11

A bag contains 10 marbles: 2 are green, 6 are red, and 2 are blue. Jessica chooses a marble at random, and without putting it ba

ck, chooses another one at random. What is the probability that the first marble is green and the second is red? Write your answer as a fraction in simplest form.
Mathematics
1 answer:
Alexxandr [17]3 years ago
3 0
P(green)=2/10=1/5
P(red, no replacement)=6/9=2/3

So, 1/5(2/3)=2/15
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Tristen is packing lunch boxes for the school trip. Each lunch box consists of 1 sandwich, 1 fruit, and 1 drink. Tristen can cho
Lerok [7]

Step-by-step explanation:

3*4*2=24

Even if he doesn't use them all, he has 24 choices.

These kind of problems are permutation type problem. Order is not critical. Imagine yourself preparing such a meal. You have choices to be made in Sandwiches, fruits and drinks. It does not matter which order you set up the lunch box. It only matters that you chose.

You can't leave one out, nor does the problem allow you to trade (say) 2 drinks for one fruit.

6 0
3 years ago
Please help <br> Find the value of x
jeka57 [31]

Answer:

54

Step-by-step explanation:

The sum of given angles must be equal to 360°

3x° + 2x° + 90° = 360°

5x + 90 = 360

5x = 270 divide both sides by 5

x = 54

3 0
3 years ago
Using the graph below, what would the input (x-value) need to be for an output (y value) of 1?
sergij07 [2.7K]

Answer:

X = 2

Y = 1

The graph intersects at the exact point of (2,1)

5 0
3 years ago
Read 2 more answers
Suppose you want to create a set of weights so that any object with an integer weight from 1 to 40 pounds can be balanced on a t
Archy [21]

Answer:

You need 4 weights, and the weights are 1; 3; 9 and 27.

Step-by-step explanation:

<em>Since the scale has two plates, we can place weights on either side and also the object so it can be balanced. </em>

This is a key part of the problem, it's not only on the other side of the scale, but on both sides.

Let's do the math now.

If i get two weights, 1 and 3. I can form this combinations.

Object of 1lb = 1

Object of 2lb + 1 weight = 3 weight.

Object of 3lb = 3 weight

Object of 4lb = 1 weight + 3 weight.

So what if i want to add the next weight and that weight to add me the maximum amount of objects. The weight would have to have a difference with the last object plus one. So if i grab 9. 9 minus 4 is 5. And that is a difference with the last object plus 1.

With a weight of 9, now i can add all the integers up to 13lb.

And the next step? Lets add one more. Keeping the last rule, the weight would have to have a difference with the last object plus one. So if i grab 27, 27 minus 13 is 14. And that is a difference witht the last object plus 1.

The sum of all the weights adds up to 40 pounds. And i can balance any integer in the middle.

The formula we are using is p – n = n + 1

Where p is the new weight. and n is the last object we weighted. And the sum of the weights goes up to the last object we can place on the scale, and in this case is 40.

4 0
3 years ago
Can anyone fill these out?
Tcecarenko [31]

Answer:

Part (a)

36

Part (b)

Find the filled table in the attachment

Part (c)

1/36

Step-by-step explanation:

The total number of possible outcomes using the multiplicative rule is given by;

6*6 = 36.

There are 6 possible outcomes in rolling each die, we simply find the product.

The probability of rolling double sixes is given by;

pr(6 and 6) = pr(6) * pr(6) = 1/6 * 1/6 = 1/36

The probability of rolling double sixes represents independent events and thus we employ the multiplicative rule of probability.

8 0
3 years ago
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