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kogti [31]
2 years ago
10

Jessie is going to the county fair. The cost of admission is $5.00 and it is $2.00 for each ride. If Jessie has $19.00, Write an

equation to determine how many rides Jessie can ride
Mathematics
2 answers:
wel2 years ago
7 0
R=7 because $2.00(7)+$5.00=$19.00
$14.00+$5.00=$19.00 so therefore she can only ride 7 rides.
AfilCa [17]2 years ago
3 0

Answer:

Step-by-step explanation:

Lets say each ride is r

2.00r + 5.00 = 19.00

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Which is more, 7,037 yards or 4 miles
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4 miles are about 7040 yards so they are very close but a mile is longer.
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Giselle made 24 1/8 ounces of lemonade. She sampled 1 1/2 ounces to make sure it is not too sour. How much lemonade is left? Exp
vivado [14]

step.1 turn both of the numbers into fraction

24x8=192, because 8 out of 8 is 1. Plus one equals to 193/8

1x2=2 plus 1 equals to 3. 3/2

step.2 turn the bottom numver the same

because 2x4 is 8 from the bottom

3x4 is 12 so the lemonade she sampled is 12/8

step.3 substract

193/8-12/8you keep the bottom the same substract the top.

193-12

the answer is : 181/8

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3 years ago
Heather made a total of $451.78 last month at her part-time job. If she worked 14 hours, how much did she make per hour?
Step2247 [10]
The amount of money divided by the amount of hours will give you how much money she makes per hour.

451.78/14=32.27.

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3 0
3 years ago
Suppose f(x) =6x second power-4, what is f(-1)?
meriva

Answer:

f(-1)=6×(-1)^-4=6/(-1)^4=6/1=6

8 0
3 years ago
Read 2 more answers
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
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