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Brilliant_brown [7]

3 years ago

9

Sam has 8 quarters, 5 dimes, and 3 Nicole's in his pocket. If one coin is selected random, what is the probability that it will

not be a quarter

1 answer:

Murrr4er [49]3 years ago

6 0

There are a total of 16 coins, so out of the ratio, that 16 goes on the bottom. Now you have to add up the coins that are NOT quarters and that number goes on the top. So your ratio is 8/16 or 1/2

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A landscaping company is laying a triangular flower bed. The height of the trangle is 15ft. What lengths of the base will make t

mash [69]

Answer:

28 ft

Step-by-step explanation:

The height of the triangle = 15 ft

The least required area A = 210 ft^2

We know that the area of a triangle = 1/2bh

Where

b is the base

h is the height

Substituting, we have

210 = 1/2 x b x 15

210 = 15/2 x b

420/15 = b

b = 28 ft

5 0

3 years ago

Find the mean:<br> 28, 18, 19, 18, 17<br> Heh help

kap26 [50]

Answer:

Mean = 20

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Perform the indicated operation & simplify. Express the answer as a complex number. ( 6 − 7 i ) ( − 8 + 3 i ) =___

slega [8]

( 6 − 7 i ) ( − 8 + 3 i )

First FOIL:

( 6 − 7 i ) ( − 8 + 3 i ) = -48 + 18i + 56i -21i^2

Combine like terms

= -48 + 74i -21i^2

Convert i^2 = -1

= -48 + 74i -21(-1)

And simplify

= -48 + 74i + 21

= -27 + 74i

6 0

2 years ago

jason mixed 8ml of a salt solution with 11ml of a 47% salt solution to make a 43%salt solution. find the percent salt concentrat

choli [55]

Let x% be the salt concentration of first solution.

Then in 8ml, amount of salt = $\frac{8x}{100}$

Given that to this 8ml solution, 11ml solution of a 47% salt solution is mixed.

amount of salt in 11ml of 47% salt solution is $\frac{11*47}{100}$ = $\frac{517}{100}$

So, total amount of salt in new mixture = $\frac{8x}{100} +\frac{517}{100} = \frac{8x+517}{100}$

But given that this new solution is a 43% salt solution.

Hence amount of salt in new mixture of (8+11)=19ml solution of 43% salt solution is $\frac{19*43}{100} = \frac{817}{100}$

Hence $\frac{8x+517}{100} = \frac{817}{100}$

8x+517 = 817

8x=817-517 = 300

x= 300/8 = 37.5

Hence the salt concentration in the first solution is 37.5%

8 0

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

3 years ago