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3 years ago

9

A bag contains 11 red and 8 blue marbles. two marbles are selected at random without replacement from the bag. what is the proba

bility that the marbles are the same color?

1 answer:

butalik [34]3 years ago

5 0

There is a 57% chance that both marbles will be red, and a 42% chance that the two marbles will be blue.

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Find the probability that when a fair coin is flipped four times at least 2 heads

salantis [7]

Since the problem states "at least" we need to also find probability of 3 H or 4 H or 5 H

Now find the probability of flipping a head 4 times; $\frac{1}{2}$ ⁴
= (1/16)

Now probability of flipping a head 3 times: (4C3)(1/2)⁴ = 4/16

Probability of flipping a head 2 times; (4C2)(1/2)⁴=6/16

(1/16)+(4/16)+(6/16)=11/16

Probability of flipping a fair coin 4 times with at least 2 heads is 11/16.

Hope I helped :)

8 0

2 years ago

Give me authentic answers please.

nataly862011 [7]

B - the value of x is positive and the value of y is negative

5 0

2 years ago

I will give you 50 if you helped me thanks

Nimfa-mama [501]

Answer:

it's 40!

10 x 10 = 100

10 x 4 = 40

6 x 10 = 60

and 6 x 4 = 24 !!

5 0

2 years ago

Read 2 more answers

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by

11111nata11111 [884]

Answer:

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is $V=l\times b\times h$

$V=(11-2x)\times (7-2x)\times x$

$V=4x^3-36x^2+77x$

Derivate w.r.t x,

$V'(x)=4(3x^2)-2(36x)+77$

$V'(x)=12x^2-72x+77$

The critical point when V'(x)=0

$12x^2-72x+77=0$

Solve by quadratic formula,

$x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}$

$x=4.607,1.392$

Derivate again w.r.t x,

$V''(x)=24x-72$

Now, $V''(4.607)=24(4.607)-72=38.568>0$ (+ve)

$V''(1.392)=24(1.392)-72=-38.592$ (-ve)

So, there is maximum at x=1.392.

The length of the box is $l=11-2x$

$l=11-2(1.392)=8.216$

The breadth of the box is $b=7-2x$

$b=7-2(1.392)=4.216$

The height of the box is h=1.392.

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is $V=l\times b\times h$

$V=8.216\times 4.216\times 1.392$

$V=48.217\ in.^3$

The volume of the box is 8.217 cubic inches.

5 0

2 years ago

A cookie factory uses 7/10 of a barrel of oatmeal in each batch of cookies. The factory used 4 9/10 barrels of oatmeal yesterday

melamori03 [73]

Answer: 7 BATCHES

Step-by-step explanation: to produce one batch it takes 7/10 of oatmeal

Now the factory uses 4 9/10 batch of oatmeal

1=7/10 OATMEAL

X=4 9/10 OATMEAL

X=(7/10*1)/4 9/10

=4 9/10*10/7

=490/70

=7

5 0

2 years ago