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Irina18 [472]
3 years ago
7

Deshawn has two bags of marbles. The first bag has 2 blue, 3 orange, and 5 red. The second bag has 4 pink, 10 blue, and 6 brown.

If he pulls a marble out of each bag, what is the probability that he will pull out a blue marble from each bag
Mathematics
1 answer:
madreJ [45]3 years ago
3 0

Answer: 0.1

Step-by-step explanation:

Given

Bag-I has 2 blue,3 orange, 5 red

Bag-II has 4 Pink,10 blue, 6 brown

No of ways of choosing a blue marble from bag-I

\Rightarrow ^2C_1

Total no of ways of choosing a marble from bag-I

\Rightarrow ^{10}C_1

No of ways of choosing a blue marble from bag-II

\Rightarrow ^{10}C_1

Total no of ways of choosing a marble from bag-II

\Rightarrow ^{20}C_1

The probability that he will pull out a blue marble from each bag is

\Rightarrow P=\text{Probability of pulling a blue marble from bag-I}\times \text{Probability of pulling a blue out bag-II}

\Rightarrow P=\dfrac{^2C_1}{^{10}C_1}\times \dfrac{^{10}C_1}{^{20}C_1}\\\\\Rightarrow P=\dfrac{2}{10}\times \dfrac{10}{20}=\dfrac{1}{10}

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Finger [1]

Given:

In triangle ABC, AB = AC, AD is angle bisector and measure of angle C is 49 degrees.

To find:

The value of x and y.

Solution:

In triangle ABC,

AB=AC                    (Given)

So, triangle ABC is an isosceles triangle and by the definition of base angles the base angles of isosceles triangle are congruent.

In isosceles triangle ABC,

\angle B\cong \angle C

m\angle B\cong m\angle C

m\angle B\cong 49^\circ

The angle bisector of an isosceles triangle is the median and altitude of the triangle. So, the angle bisector is perpendicular to the base.

m\angle ADB=90^\circ

x^\circ=90^\circ

In triangle ABD,

m\angle DAB+m\angle ABD+m\angle ADB=180^\circ         [Angle sum property]

y^\circ+49^\circ+x^\circ=180^\circ

y^\circ+49^\circ+90^\circ=180^\circ

y^\circ=180^\circ-49^\circ-90^\circ

y^\circ=41^\circ

Therefore, the correct option is B.

3 0
3 years ago
Given f(x) and g(x) = f(x) + k, look at the graph below and determine the value of k.<br> k=______.
LenKa [72]

Answer:

5

Step-by-step explanation:

Look at one of the coordinates to find k. For this example I'll look at the y-intercepts (0, 3) (for g(x)) and (0, -2) (for f(x))

g(0) = 3

g(0) = f(0) + k

3 = -2 + k

5 = k

7 0
2 years ago
Ajay is on the swim team. He improved his time on
ozzi

Answer:

<u>5%</u>

Step-by-step explanation

50/32= 1.56yard per second        ; 50/31= 1.61 yard per second

1.61-1.56= 0.05* 100= 5%

5 0
3 years ago
If the cube shown above is sliced by a plane to create a rectangle that is not a square, which sets of vertices could the plane
Veronika [31]
If I’m sure it should be D or C
3 0
2 years ago
In a bag of m&amp;m's there are 5 brown 6 yellow 4 blue 3 green and 2 orange. What's the probability of getting 3 yellow m&amp;m
olasank [31]
There are 5+6+4+3+2=20 m&m's in the bag.
Calculate in how many ways you can choose 3 m&m's from 20:
_{20} C _3=\frac{20!}{3!(20-3)!}=\frac{20!}{3! \times 17!}=\frac{17! \times 18 \times 19 \times 20}{6 \times 17!}=\frac{18 \times 19 \times 20}{6}=3 \times 19 \times 20= \\&#10;=1140

There are 6 yellow m&m's.
Calculate in how many ways you can choose 3 m&m's from 6:
_6 C _3 = \frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=\frac{4 \times 5 \times 6}{6}=4 \times 5=20

The probability is the number of ways of choosing 3 m&m's from 6 m&m's divided by the number of ways of choosing 3 m&m's from 20 m&m's.
P=&#10;\frac{20}{1140}=\frac{20 \div 20}{1140 \div 20}=\frac{1}{57}

The probability is 1/57.
4 0
3 years ago
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