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chubhunter [2.5K]
3 years ago
8

There are 12 pieces of fruit in a bowl five are lemons and the rest are limes. You choose a piece of fruit without looking. The

piece of fruit is a lime
Mathematics
1 answer:
Sati [7]3 years ago
5 0

I agree. So like what you already said, there's 5 out of 12 chance that you will get lemon. And 7 out of 12 chances, you will get lime.

Ratio Format:

5:12 chances for lemon

7:12 chances for lime

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In 2010, there were about 246 million vehicles (cars and trucks) and about 308.7 million people in the United States.1 The numbe
belka [17]

Answer:

2054

Step-by-step explanation:

The equation for the growth of people = 308.7(1097)^x

The equation for the growth of vehicles = 246(1155)^x

If the number of both are equal tham

308.7(1097)^x=246(1155)^x

Dividing both sides by 246

1.254878(1097)^x=(1155)^x

Dividing both sides by (1097)^x

1.254878=(1155÷1097)^x

X=4.406 decades

X=44 years

5 0
3 years ago
Which equation is true when n=1.2
enot [183]

Answer:

5+1.2=6.2

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
What is 18 divided by negative 3
sweet-ann [11.9K]
18 / -3 = -6

answer

<span>negative 6 is 18 divided by negative 3
</span>
hope it helps
6 0
4 years ago
Read 2 more answers
Y=2 Enter the solution (x, y) to the system of equations shown 7y+3x=5
AnnyKZ [126]

Answer:

(x, y) = (-3, 2)

Step-by-step explanation:

Put the given value of y into the equation and solve for x.

... 7·2 +3x = 5

... 3x = -9 . . . . . . subtract 14

... x = -3 . . . . . . . divide by the x-coefficient

The solution (x, y) is (-3, 2).

6 0
3 years ago
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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