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Lady bird [3.3K]
3 years ago
15

Can someone simplify this please :) 6.89655172414

Mathematics
1 answer:
natulia [17]3 years ago
3 0
You can’t simplify that
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Caitlin went to the store to buy school clothes .She had a store credit from from a previous return in the amount of $39.58 if s
ZanzabumX [31]
52.22
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13.64
÷ 4
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3.41
3 0
3 years ago
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A ribbon is 5 2 5 yards long. Mae needs to cut the ribbon into pieces that are 2 5 of a yard long. How many pieces of ribbon can
Rom4ik [11]

Answer: The number of pieces of the ribbon she can cut = 13\frac12 yards

Step-by-step explanation:

Given: The length of the ribbon = 5\dfrac25 yards

= \dfrac{5\times5+2}{5}  =\dfrac{27}{5} yards

Length of each piece = \dfrac25 yards

The number of identical pieces can be cut from the entire ribbon = (Length of the ribbon ) ÷ (Length of each piece )

=\dfrac{27}{5}\div\dfrac{2}{5}\\\\=\dfrac{27}{5}\times\dfrac{5}{2}\\\\=\dfrac{27}{2}\\\\=13\dfrac12

Hence, the number of pieces of the ribbon she can cut = 13\frac12 yards

7 0
3 years ago
The game of clue involves 6 suspects, 6 weapons, and 9 rooms. one of each is randomly chosen and the object of the game is to gu
mina [271]
Part A:

Given that t<span>he game of clue involves 6 suspects, 6 weapons, and 9 rooms.

The number of ways that one of each is randomly chosen is given by:

^6C_1\times{ ^6C_1}\times{ ^9C_1}=6\times6\times9=324

Therefore, the number of solutions possible is 324.



Part B:

Given that a </span>players is randomly given three of the remaining cards, <span>let s, w, and r be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player.

The number of suspects, weapons, and rooms remaining respectively after the player observes his or her three cards are: 6 - s, 6 - w, and 9 - r.

Let x denote the number of solutions that are possible after that player observes his or her three cards, then:

x={ ^{6-s}C_1}\times{ ^{6-w}C_1}\times{ ^{9-r}C_1}=(6-s)(6-w)(9-r)

Therefore, x in terms of s, w, and r is given by x = (6 - s)(6 - w)(9 - r).



Part C:

The expected value E(x) of a data set x_i with probabilities p(x_i) is given by E(x)=\Sigma xp(x)

There are </span>^{3+3-1}C_{3-1}={ ^5C_2}=10 possible combinations s, w and r. They are (3, 0, 0), (0, 3, 0), (0, 0, 3), (2, 1, 0), (0, 2, 1), (1, 0, 2), (2, 0, 1), (1, 2, 0), (0, 1, 2), (1, 1, 1)

Thus the expected value is given by

E(x)=3\cdot6\cdot9p(3, 0, 0)+6\cdot3\cdot9p(0, 3, 0)+6\cdot6\cdot6p(0, 0, 3) \\ 4\cdot5\cdot9p(2, 1, 0)+6\cdot4\cdot8p(0, 2, 1)+5\cdot6\cdot7p(1, 0, 2)+4\cdot6\cdot8p(2, 0, 1) \\ +5\cdot4\cdot9p(1, 2, 0)+6\cdot5\cdot7p(0, 1, 2)+5\cdot5\cdot8(1, 1, 1) \\  \\ = \frac{1}{ ^{21}C_3} (162\cdot{ ^6C_3}\cdot{ ^6C_0}\cdot{ ^9C_0}+162\cdot{ ^6C_0}\cdot{ ^6C_3}\cdot{ ^9C_0}+216\cdot{ ^6C_0}\cdot{ ^6C_0}\cdot{ ^9C_3} \\ \\ +180\cdot{ ^6C_2}\cdot{ ^6C_1}\cdot{ ^9C_0}+192\cdot{ ^6C_0}\cdot{ ^6C_2}\cdot{ ^9C_1}

+210\cdot{ ^6C_1}\cdot{ ^6C_0}\cdot{ ^9C_2}+192\cdot{ ^6C_2}\cdot{ ^6C_0}\cdot{ ^9C_1}+180\cdot{ ^6C_1}\cdot{ ^6C_2}\cdot{ ^9C_0} \\  \\ +210\cdot{ ^6C_0}\cdot{ ^6C_1}\cdot{ ^9C_2}+200\cdot{ ^6C_1}\cdot{ ^6C_1}\cdot{ ^9C_1} \\  \\ =\frac{1}{1,330}(324\cdot20+216\cdot84+360\cdot90+384\cdot135+420\cdot216+200\cdot324) \\  \\ =\frac{1}{1,330}(6,480+18,144+32,400+51,840+90,720+64,800) \\  \\ =\frac{1}{1,330}(264,384) \\  \\ =\bold{198.78}
6 0
3 years ago
Julian’s friend James also has a tank in the shape of a triangular prism. The dimensions of James’ tank are each exactly 2 3/4 t
malfutka [58]

The approximate area of the triangular base of James’ tank is 3025 in²

To find the approximate area of the triangular base of James' tank, we need to know what scale is.

<h3>What is scale?</h3>

A scale is the ratio of two dimension. It is given by scale = New dimension/old dimension

Given that the dimensions of James’ tank are each exactly 2 3/4 times the dimensions of Julian’s tank.

The scale of the area of James' tank to the area of triangular base of Julian's tank is S = (2³/₄)²

= (11/4)²

= 121/16

Let

  • A = area of triangular base of James' tank,
  • A' = area of triangular base of Julian's tank = 400 in²

So, scale, S = A/A'

<h3>The approximate area of the triangular base of James’ tank</h3>

So, making A subject of the formula, we have

A = SA'

A = 121/16 × 400 in²

A = 121 × 25 in²

A = 3025 in²

So, the approximate area of the triangular base of James’ tank is 3025 in²

Learn more about area of triangle here:

brainly.com/question/27436493

#SPJ1

5 0
3 years ago
What is the answer for x+11+8x=29 has to show work
m_a_m_a [10]

Answer:

x=2

Step-by-step explanation:

9x+11=29

9x=29-11

9x=18

x=2

8 0
3 years ago
Read 2 more answers
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