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Lubov Fominskaja [6]
3 years ago
5

The customer states that she was shorted $0.23 change from $3 instead of $0.32 how much the holler back

Mathematics
1 answer:
muminat3 years ago
6 0

Answer:

9 cents?

Step-by-step explanation:

she has $3 she was supposed to get back 32 cents but ended up getting 23 back. so there for 9 cents

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2(x−3)=8(x−6) what is x?
uysha [10]

Answer: x = 7

Step-by-step explanation:

2x - 6 = 8x - 48

+48

2x + 42 = 8x

-2x

42 = 6x

/6

7 = x

4 0
3 years ago
Read 2 more answers
Perform the following translations on the giving figures.
Kazeer [188]

Answer:

(-3, +4)Q

(+4, +4)R

(+3, -4)T

(-4, -4)S

3 0
2 years ago
Every sixth visitor to an animal gets a free animal calendar.every twentieth visitor gets a free animal toy. which visitor each
blondinia [14]
The 60th customer each day will get both a calendar and a toy, as 60/20 = 3 and 60/6 = 10
8 0
3 years ago
In a large school, it was found that 71% of students are taking a math class, 77% of student are taking an English class, and 58
Anni [7]

Answer:

P (Math or English) = 0.90

Step-by-step explanation:

* Lets study the meaning of or , and on probability

- The use of the word or means that you are calculating the probability

  that either event A or event B happened

- Both events do not have to happen

- The use of the word and, means that both event A and B have to

  happened

* The addition rules are:

# P(A or B) = P(A) + P(B) ⇒ mutually exclusive (events cannot happen

  at the same time)

# P(A or B) = P(A) + P(B) - P(A and B) ⇒ non-mutually exclusive (if they  

   have at least one outcome in common)

- The union is written as A ∪ B or “A or B”.  

- The Both is written as A ∩ B or “A and B”

* Lets solve the question

- The probability of taking Math class 71%

- The probability of taking English class 77%

- The probability of taking both classes is 58%

∵ P(Math) = 71% = 0.71

∵ P(English) = 77% = 0.77

∵ P(Math and English) = 58% = 0.58

- To find P(Math or English) use the rule of non-mutually exclusive

∵ P(A or B) = P(A) + P(B) - P(A and B)

∴ P(Math or English) = P(Math) + P(English) - P(Math and English)

- Lets substitute the values of P(Math) , P(English) , P(Math and English)

 in the rule

∵ P(Math or English) = 0.71 + 0.77 - 0.58 ⇒ simplify

∴ P(Math or English) =  0.90

* P(Math or English) = 0.90

8 0
3 years ago
What are the answers to these 2 question 4&5 PLZ HELP
Ahat [919]
The first problem:

y = -2 x²  ⇒⇒⇒ x² = -y/2 →(1)33 x² + y² = 27 → (2)
By substitution from (1) at (2) with the value of x²
∴ 33 (-y/2) + y² = 27
∴ y² - 16.5 y - 27 = 0
a = 1 , b = -16.5 , c = -27
∴ y= \frac{-b \pm  \sqrt{b^2-4ac} }{2a} = \frac{16.5 \pm  \sqrt{(-16.5)^2-4*1*(-27)} }{2*1}
∴ y = 18 or y = -3/2
By substitution from at (1) with the value of y
for y = 18   ⇒⇒⇒ x² = -18/2 = -9 (unacceptable)
for y = -3/2 ⇒⇒⇒ x² = -(-3/2)/2 = 3/4
∴ x= \pm \sqrt{ \frac{3}{4} } = \pm  \frac{ \sqrt{3} }{2} 

The correct options are 2 , 7
Solution of the system of equations is 
( \frac{ \sqrt{3} }{2} ,  \frac{-3}{2} )
and   
( -\frac{\sqrt{3} }{2} ,  \frac{-3}{2} )
==================================
The second problem:
The general equation of the hyperbole is 
\frac{x^2}{a^2} - \frac{y^2}{b^2} =1

Transverse axis is horizontal 

The equation if the asymptotes are y = \pm \frac{b}{a}x
For the given equation:
\frac{x^2}{225} -  \frac{y^2}{36} = 1
a² = 225 ⇒⇒⇒ a = √225 = 15
b² = 36   ⇒⇒⇒ b = √36 = 6
∴ the slope of the <span><span>asymptotes = </span>b/a   and  -b/a
</span><span>
</span><span>b/a = 6/15 = 2/5
</span><span>
</span><span>-b/a = -6/15 = -2/5
</span><span>
</span><span>∴ m = 2/5   and   m = -2/5
</span>






3 0
3 years ago
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