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yKpoI14uk [10]
2 years ago
5

A bag contains 6 nickels and 14 pennies. What is the ratio, written in lowest terms, of nickels to the total number of coins in

the bag?
Mathematics
2 answers:
VLD [36.1K]2 years ago
4 0
There are 6 nickels and 14 pennies.
There are 20 total coins.

The ratio of nickels to coins is 6:20.
Both of these are divisible by 2, so we can reduce this to 3:10.
Zigmanuir [339]2 years ago
3 0
A ratio is written like it says like if there are 2 pennies and 5 quarters, if you were to make a ratio out of that, then the ratio is 2:5
The amount of nickels is 6 and the total number of coins is 20
So this ratio would be 6:20 or 6/20 or 6 to 20, however you want to write it
When the problem asks the lowest terms, it means the simplest ratio where you cannot simplify the ratio any further like 1:2 is simplified but 2:8 isn't simplified because you can reduce 2:8 into 1:4
First, you must find the GCF (otherwised known as greatest common factor)
This can be confused with LCM
The GCF is no more than the smallest number
The LCM is no less than your greatest number
If you can't get the GCF of a ratio, then that ratio is simplified and can no longer be simplified.
So the GCF of 6 and 20 is 2
You need to divide both numbers by the GCF and you will get
3:10
<em>The ratio from nickels to pennies is 3 nickels / 10 coins</em>
This is how I learned it in school and I know some methods are probably easier than this one.
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0.6x3.2

Step-by-step explanation:

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Which of the slopes below represents a line that is perpendicular to the line =13+4 ?
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3 years ago
Bob and Anna are planning to meet for lunch at Sally's Restaurant, but they forgot to schedule a time. Bob and Anna are each goi
Viktor [21]

Answer:

The expected cost is $8.75

<em></em>

Step-by-step explanation:

Given

Time = \{1pm, 2pm, 3pm, 4pm\}

C_1 = \$5 --- If Bob and Anna meet

C_2 = \$10 --- If Bob and Anna do not meet

Required

The expected cost of  Bob's meal

First, we list out all possible time both Bob and Anna can select

We have:

(Bob,Anna) = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)\}

n(Bob, Anna) = 16

The outcome of them meeting at the same time is:

Same\ Time = \{(1,1),(2,2),(3,3),(4,4)\}

n(Same\ Time) = 4

The probability of them meeting at the same time is:

Pr(Same\ Time)  = \frac{n(Same\ Time)}{n(Bob,Anna)}

Pr(Same\ Time)  = \frac{4}{16}

Pr(Same\ Time)  = \frac{1}{4}

The outcome of them not meeting:

Different = \(Bob,Anna) = \{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2)

,(3,4),(4,1),(4,2),(4,3)\}

n(Different) = 12

The probability of them meeting at the same time is:

Pr(Different)  = \frac{n(Different)}{n(Bob,Anna)}

Pr(Different)  = \frac{12}{16}

Pr(Different)  = \frac{3}{4}

The expected cost is then calculated as:

Expected = C_1 * P(Same) + C_2 * P(Different)

Expected = \$5 * \frac{1}{4} + \$10 * \frac{3}{4}

Expected = \frac{\$5}{4} + \frac{\$30}{4}

Take LCM

Expected = \frac{\$5+\$30}{4}

Expected = \frac{\$35}{4}

Expected = \$8.75

<em>The expected cost is $8.75</em>

4 0
2 years ago
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maksim [4K]

Answer:

i believe it is a if not im sorry

Step-by-step explanation:

5 0
3 years ago
Show that there do not exist scalar c1 c2 and c3 c1(-2,9,6)+c2(-3,2,1)+c3(1,7,5)= (0,5,4)​
fredd [130]

This is the same as showing the following system of equations doesn't have a solution:

\begin{cases}-2c_1-3c_2+c_3 = 0 \\ 9c_1 + 2c_2 + 7c_3 = 5 \\ 6c_1 + c_2 + 5c_3 = 4\end{cases}

or in matrix form,

\begin{bmatrix}-2&9&6\\-3&2&1\\1&7&5\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix} = \begin{bmatrix}0\\5\\4\end{bmatrix}

The quickest way to check if there is a solution is to check whether the coefficient matrix is invertible. If its determinant is 0, then it is not invertible.

And the quickest way to show that the determinant is 0 is by observing that the third row is a linear combination of the first two rows:

(-2, 9, 6) - (-3, 2, 1) = (-2 + 3, 9 - 2, 6 - 1) = (1, 7, 5)

So there are indeed no such scalars <em>c₁</em>, <em>c₂</em>, and <em>c₃</em>.

5 0
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