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RoseWind [281]
3 years ago
10

Oki so help plz and show proof or evidence

Mathematics
1 answer:
muminat3 years ago
5 0

Answer:

Option B because the initial fee stays the same which means that there will still be + 70 and then they added 60.

60t + 70

Step-by-step explanation:

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If a 12 ounce bag of popcorn at the theater cost $5.50. How much would a 30 ounce bag cost? $10.00
Kaylis [27]

Answer:

I think it’s C I’m not sure I seen a question like this before

Step-by-step explanation:

4 0
3 years ago
The value of X in the figure ABCD witch is a rhombus
NeX [460]

Answer:

  x = 10°

Step-by-step explanation:

The diagonals of a rhombus divide the figure into congruent right triangles. That means the two marked acute angles are complementary.

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<h3>solve for x</h3>

Complementary angles total 90°, so the relation we can use to find x is ...

  (3x -13) +(8x -7) = 90

  11x = 110 . . . . . . . . . add 20

  x = 10 . . . . . . . . . divide by the coefficient of x

The value of x is 10°.

5 0
2 years ago
-6+6=0 is an example of what property ?
notka56 [123]

Step-by-step explanation:

communitive property

4 0
3 years ago
2. Which expression is equivalent to 3x(-4x² + 5x - 8) - 6(x² - 5x + 7)?
adoni [48]
It is answer choice B. Using the distributive property, -12x^3+15x^2-24x-6x^2+30x-42. Then combine like terms.
5 0
3 years ago
A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If
Blababa [14]

Answer: Our required probability is \dfrac{1}{2}

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = \dfrac{1}{3}

So, Probability of getting head from fair coin = \dfrac{1}{2}

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}

Hence, our required probability is \dfrac{1}{2}

No, the answer is not \dfrac{1}{3}

5 0
3 years ago
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