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vodomira [7]
3 years ago
12

A 10-yr-old competes in gymnastics. For several competitions, she received the following "All-Around" scores: 35.5, 36.3. 36.6,

and 36.9. Her coach recommends that gymnasts whose "All-Around" scores average at least 36.5 moves up to the next level. What "Ail-Around" scores in the next competition would result in the child being eligible to move up? The child needs a score of _____ to move up to the next level of the competition.
Mathematics
1 answer:
Delicious77 [7]3 years ago
3 0

Answer:

The child needs a score of 37.2 to move up to the next level of the competition.

Step-by-step explanation:

The mean is the sum of all scores divided by the number of competions. So

M = \frac{S}{T}

In which S is the sum of all her scores and T is the number of competitions.

The child has five competions:

Which means that T = 5

She has to get a mean of at least 36.5, so M = 36.5

Her scores are: 35.5, 36.3. 36.6, and 36.9. Her last score, i am going to call x. So

S = 35.5 + 36.3 + 36.6 + 36.9 + x = 145.3 + x

The child needs a score of _____ to move up to the next level of the competition.

This score is x. So

M = \frac{S}{T}

36.5 = \frac{145.3 + x}{5}

145.3 + x = 36.5*5

x = 37.2

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2. Solve the system of equations. Write your answer in parametric vector notation. X, +5x2 + 4x3 + 3x4 +9xs = 18 2x, +6x2 + 4x3
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Answer:

The system is inconsistent.

Step-by-step explanation:

The system is

x_1+5x_2+4x_3+3x_4+9x_5=18\\2x_1+6x_2+4x_3+6x_4+6x_5=28\\3x_1+7x_2+4x_3+10x_4+x_5=41\\4x_1+6x_2+2x_3+7x_4+4x_5=29

The associated matrix to the system is

\left[\begin{array}{cccccc}1&5&4&3&9&18\\2&6&4&6&6&28\\3&7&4&10&1&41\\4&6&2&7&4&29\end{array}\right]

Now we use row operations to find the echelon form of the matrix:

1. We substract to row 2, two times the row 1.

We substract to row 3, three times the row 1.

We substract to row 4, four times the row 1 and obtain the matrix

\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&-4&-4&0&-12&-8\\0&-8&-8&1&-26&-13\\0&-14&-14&-5&-32&-43\end{array}\right]

2. We multiply the second row of the preview step by -1/4. We obtain the matrix

\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&-8&-8&1&-26&-13\\0&-14&-14&-5&-32&-43\end{array}\right]

3.

We add to row 3, eight times the row 2.

We add to row 4, fourtheen times the row 2 and obtain the matrix

\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&0&0&1&-2&3\\0&0&0&-5&10&-155\end{array}\right]

4. We add to row 4 of the preview step, five times the row 3 and obtain the matrix

\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&0&0&1&-2&3\\0&0&0&0&0&-140\end{array}\right]

Using backward substitution we have that

0x_5=-140, then 0=-140 and this is absurd. Then The system is inconsistent.

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