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3 years ago

14

There are 3 white marbles and 7 blue marbles in a bag. Jamie will randomly pick two marbles out of the bag without replacing the

first marble chosen. What is the probability of Jamie's picking 2 white marbles?
A) 1/15

B) 1/18

C) 1/20

D) 1/7

2 answers:

Softa [21]3 years ago

7 0

Answer:

A) 1/15

Step-by-step explanation:

The probability of picking a white marble is 3/10. Then 2 of the remaining 9 marbles are white, so the probability of picking one is 2/9.

The joint probability is (3/10)·(2/9) = 6/90 = 1/15

ELEN [110]3 years ago

5 0

Answer:

1/7 is the probability hope this helps

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3 years ago

16 divided by 6.08 .<br> 2 digit divisors

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3 years ago

Read 2 more answers

Are these answers right

bixtya [17]

yes they are and C has no answer in it and can i have branilest

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3 years ago

I need help with this one to

Aleks [24]

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3 years ago

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by

a_sh-v [17]

Answer:

Dimension of the box is $16.1\times 7.1\times 2.45$

The volume of the box is 280.05 in³.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

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

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is $L=21-2h$

The width of the box is $W=12-2h$

The volume of the box is $V=L\times W\times H$

$V=(21-2h)\times (12-2h)\times h$

$V=(21-2h)\times (12h-2h^2)$

$V=252h-42h^2-24h^2+4h^3$

$V=4h^3-66h^2+252h$

To maximize the volume we find derivative of volume and put it to zero.

$V'=12h^2-132h+252$

$0=12h^2-132h+252$

Solving by quadratic formula,

$h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}$

$h=\frac{132\pm72.99}{24}$

$h=2.45,8.54$

Now, substitute the value of h in the volume,

$V=4h^3-66h^2+252h$

When, h=2.45

$V=4(2.45)^3-66(2.45)^2+252(2.45)$

$V\approx 280.05$

When, h=8.54

$V=4(8.54)^3-66(8.54)^2+252(8.54)$

$V\approx -170.06$

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is $L=21-2(2.45)=16.1$

The width of the box is $W=12-2(2.45)=7.1$

So, Dimension of the box is $16.1\times 7.1\times 2.45$

6 0

4 years ago