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4vir4ik [10]
3 years ago
10

The first class had a kids in it, the second had b kids in it, and the third class had c kids in it. Kids from all three classes

were equally divided between two buses. How many kids were on each bus?
Mathematics
2 answers:
astra-53 [7]3 years ago
5 0
There are no numbers and only variables so I can't tell you the amount of kids on each bus but here is the equation:
(a*b*c)/2
kramer3 years ago
3 0
(a+b+c)/2. this is your answer. if it had the numbers i would solve it fully, but it only has variables.
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A total of
Leviafan [203]
Total tickets sold = 417

2 fractions goes to students.
1 fraction goes to adults.

Total fractions of tickets
= 2 + 1
= 3 fractions

Since adults = 1 fraction
1 fraction = 417 ÷ 3
= 139 tickets

3 0
3 years ago
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Which of the following is the reciprocal parent function A.f(x)= -x^2 B.F(x)= x+1 C. F(x)=|x| D.F(x)=1/x
Cerrena [4.2K]
The reciprocal parent function would be, F(x)= 1/x
6 0
3 years ago
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A= [1, 3; 2, 1], B=[3, 6; -1, 1]. Find AB & BA if possible
steposvetlana [31]

Answer:

AB\Rightarrow \quad \begin{bmatrix}0 & 9\\ 5 & 13\end{bmatrix}\\BA\Rightarrow \quad \begin{bmatrix}15 & 15\\ 1 & -2\end{bmatrix}

Step-by-step explanation:

For two matrix P and Q, the product, say PQ is defined when:

The number of columns of P = The number of rows of Q

Since A is a 2×2 matrix and B is also a 2×2 matrix

Thus both AB and BA are possible.

So AB is:

AB\Rightarrow\begin{bmatrix}1 & 3\\ 2 & 1\end{bmatrix}\begin{bmatrix}3 & 6\\ -1 & 1\end{bmatrix}\\AB\Rightarrow\quad \begin{bmatrix}3\times 1+3\times (-1) & 6\times 1+3\times 1\\3\times 2+1\times (-1) & 6\times 2+1\times 1\end{bmatrix}\\AB\Rightarrow \quad \begin{bmatrix}0 & 9\\ 5 & 13\end{bmatrix}

BA is:

BA\Rightarrow\begin{bmatrix}3 & 6\\ -1 & 1\end{bmatrix}\begin{bmatrix}1 & 3\\ 2 & 1\end{bmatrix}\\BA\Rightarrow\quad \begin{bmatrix}3\times 1+6\times 2 & 3\times 3+6\times 1\\(-1)\times 1+1\times 2 & (-1)\times 3+1\times 1\end{bmatrix}\\BA\Rightarrow \quad \begin{bmatrix}15 & 15\\ 1 & -2\end{bmatrix}

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=5%5E%7B-3%7D" id="TexFormula1" title="5^{-3}" alt="5^{-3}" align="absmiddle" class="latex-form
AveGali [126]

Answer:

the equivalent equation of

5^{-3}=\frac{1}{5^{3}}=\frac{1}{125}

5 0
2 years ago
When Carson runs the 400 meter dash, his finishing times are normally distributed with a mean of 63 seconds and a standard devia
Gnom [1K]

Answer: in 95% of races, his finishing time will be between 62 and 64 seconds.

Step-by-step explanation:

The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean . The empirical rule is further illustrated below

68% of data falls within the first standard deviation from the mean.

95% fall within two standard deviations.

99.7% fall within three standard deviations.

From the information given, the mean is 63 seconds and the standard deviation is 5 seconds.

2 standard deviations = 2 × 0.5 = 1

63 - 1 = 62 seconds

63 + 1 = 64 seconds

Therefore, in 95% of races, his finishing time will be between 62 and 64 seconds.

6 0
2 years ago
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