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Hatshy [7]
3 years ago
7

A theater group made appearances in two cities. The hotel charge before tax in the second city was $ 500 lower than in the first

. The tax in the first city was 3.5 % , and the tax in the second city was 7.5 % . The total hotel tax paid for the two cities was $ 540 . How much was the hotel charge in each city before tax?
Mathematics
1 answer:
xxTIMURxx [149]3 years ago
7 0
It would have to asmbother people. It would be 12.3%
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adoni [48]
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Write two others names for fg
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Answer: FD, GF
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3 years ago
Subtract 6 from me then multiply by 2 if you subtract 40 then divide by 4 you get 8 what number I am
stellarik [79]

Answer:

42

Step-by-step explanation:

42-6 = 36

36*2 = 72

72-40 = 32

32/4 = 8

8 0
2 years ago
2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):
zlopas [31]
<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)

First pair of opposite sides:

<u>Slope:</u>

\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}

Second pair of opposite sides:

<u>Slope:</u>

\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\

So the diagonals measure the same, therefore this is a rectangle.

5 0
3 years ago
Carolina goes to a paintball field that charges an entrance fee of \$18$18dollar sign, 18 and \$0.08$0.08dollar sign, 0, point,
FromTheMoon [43]

Answer: At least 713 paintballs

Step-by-step explanation:

Based on the information that has been given in the question, the inequality that'll be used to solve the question will be:

18+0.08B≥75

0.08B≥75-18

0.08B≥57

B≥57/0.08

B≥712.5

Carolina needs to buy at least 713 paintballs along with the entrance fee to get the promotion.

4 0
3 years ago
Read 2 more answers
In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any
Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

The second footballer can be born on any other day, so he has 364 possible birthdays:

d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}

the probability for the first two footballers to be born on two different days is thus

1 \cdot \dfrac{364}{365} = \dfrac{364}{365}

Similarly, the third footballer can be born on any day, except d_1 and d_2:

d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}

so, the probability for the first three footballers to be born on three different days is

1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

So, the probability that at least two footballers are born on the same day is

1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

8 0
3 years ago
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