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11Alexandr11 [23.1K]
2 years ago
12

Pay (dollars)

Mathematics
1 answer:
Kazeer [188]2 years ago
3 0

Answer:

2 dollars per hour

Step-by-step explanation:

I believe your supposed to divide 20 by 10 to get 2.

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The polygon below are similar. solve for x and y
velikii [3]
We would need the info about the lengths of the sides of said polygon.
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Tina got a haircut, and her hair is still at least 15 inches long, write and graph an inequality​
ivanzaharov [21]

The inequality is:

x\geq15

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3 years ago
Will this happen in avenger 4?
Marizza181 [45]

In chess, the Endgame is where you sacrafice pawns, or in this case, minor characters in order to get a powerful piece back. You can leave some pawns in battle while regaining power pieces. The pawns sacraficed were Peter, Stephen, Bucky, Drax, T'challa. Mantis, etc. While leaving behind two pawns: Nebula, and Bruce Banner. Normally I like to think of the Hulk as a Rook, but since he's completely useless at the moment, he's a pawn. Nebula's fairly worthless, so she's a pawn. Thanos is playing with all power pieces and one pawn: Gamora. He sacraficed his pawn in order to complete his queen equivilence: the gauntlet. Now he's playing with all power pieces, while the Avenger's have sacraficed their pawns in order to get their queen: Captain Marvel, who in turn will wage war on Thanos only to find that a pawn has made it across the board and turned into the Hulk, and fights side by side the original Avengers to get the soul stone, revive Tony, who probably dies, get their friends home, welcome new friends, and kill Thanos.

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4 0
3 years ago
Consider rolling two fair dice and observing the number of spots on the resulting upward face of each one. Letting A be the even
Allushta [10]

Answer:

P(E|A)= \frac{10}{11}

Step-by-step explanation:

Given

Two rolls of die

E \to one of the outcomes is 6

A \to atleast one is 6

Required

P(E|A)

First, list out the outcome of each

E = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}

A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}

So:

P(E|A)= \frac{n(E\ n\ A)}{n(A)}

Where:

E\ n\ A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}

n(E\ n\ A) = 10

n(A) = 11

So:

P(E|A)= \frac{10}{11}

7 0
3 years ago
Two factory plants are making TV panels. Yesterday, Plant A produced 16,000 panels. Five percent of the panels from Plant A and
weeeeeb [17]

Answer:

Number of panels produced by Plant B is 8,000.

Step-by-step explanation:

The number of panels produced by Plant A  = 16,000

The number of defective panels from A = 5%

Now, 5% of 16,000 = \frac{5}{100}  \times 16,000 = 800

So, out of  16,000 panels produced by pant A , 800 are defective. .. (1)

Let us assume number of panels produced by Plant B  = m

The number of defective panels from B = 2%

Now, 2% of m = \frac{2}{100}  \times m = 0.02 m

So, out of total m panels produced by pant B , 0.02 m are defective. .. (2)

Now, total panels produced over all = Panels by A  + Panels by B

= 16,000 + m

The percentage of defected panels over all = 4%

Now, 4% of (16,000 + m) = \frac{4}{100}  \times (16,000+m)  = (0.04)(16,000 + m)

Also, the total number of defective panels = Defective from A  + Defective from B

⇒(0.04)(16,000 + m)  =  800 +   0.02 m   from (1) and (2)

or, 640 + 0.04 m = 800 + 0.02

or, 0.02 m = 160

⇒ m = 160 /0.02 = 8,000

or, m = 8,000

Hence, Number of panels produced by Plant B is 8,000.

5 0
3 years ago
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