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3 years ago

Keith swam 50 72 meters in the swimming pool this morning. This

Mathematics

1 answer:

ruslelena [56]3 years ago

8 0

Answer:

122.96 meters is the answer

Step-by-step explanation:

If you add you will get the answer

Hope it helpes

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The table shows the results of a survey in which 10th-grade students were asked how many siblings (brothers and/or sisters) they

lesantik [10]

Answer:

70%

Step-by-step explanation:

Given

Number of Siblings: || 0 || 1 || 2 || 3

Number of Students: || 4 || 18 || 10 || 8

Required

Determine the probability of a student having at least one but not more than 2 siblings

First, we have to determine the total number of 10th grade students

$Total = 4 + 18 + 10 + 8$

$Total = 40$

The probability of a student having at least one but not more than 2 siblings = P(1) + P(2)

Solving for P(1)

P(1) = number of students with 1 sibling / total number of students

From the given parameters, we have that:

Number of students with 1 sibling = 18

So:

$P(1) = \frac{18}{40}$

Solving for P(2)

P(2) = number of students with 2 siblings / total number of students

From the given parameters, we have that:

Number of students with 2 siblings = 10

So:

$P(2) = \frac{10}{40}$

$P(1) + P(2) = \frac{18}{40} + \frac{10}{40}$

Take LCM

$P(1) + P(2) = \frac{18 + 10}{40}$

$P(1) + P(2) = \frac{28}{40}$

Divide numerator and denominator by 4

$P(1) + P(2) = \frac{7}{10}$

$P(1) + P(2) = 0.7$

Convert to percentage

$P(1) + P(2) = 70\%$

Hence, the required probability is 70%

6 0

3 years ago

How my population be a factor in tropical rain forest biome if the area receives very little rain for a extended period of time

egoroff_w [7]

The animals will get very hot that they will start eating each other and going insane

3 0

3 years ago

Help me with this!! I dont understand this question

Thepotemich [5.8K]

Answer:

3) C

Step-by-step explanation:

The question is asking which pair is equal. Work through each problem to solve.

8(2r) and 10r

16r ≠ 10r

6r + 3 and 9r

6r + 3 ≠ 9r

9(3r - 4) and 27r - 36

27r - 36 = 27r - 36

r + r + r + r + r and r⁵

5r ≠ r⁵

7 0

4 years ago

HELPPPPPP PLEASEEEE NEED A ANSWER

son4ous [18]

Answer:

It's A

Step-by-step explanation:

5 0

3 years ago

Determine whether the given system has a unique solution, no solution, or infinitely many solutions.

victus00 [196]

Answer:

The system has no solution.

Step-by-step explanation:

To find the solution to this system of linear equations

$\left\begin{array}{ccccc}-3x_1&+x_2&-2x_3&=&8&\\x_1&+5x_2&-x_3&=&4&\\-x_1&+11x_2&-4x_3&=&1&\end{array}\right$

First, state the problem in matrix form, this means, extracting only the numbers, and putting them in a box.

$\left[ \begin{array}{ccc|c} -3 & 1 & -2 & 8 \\\\ 1 & 5 & -3 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

This is called an augmented matrix. The word “augmented” refers to the vertical line, which we draw to remind ourselves where the equals sign belong

Next, transform the augmented matrix to the reduced row echelon form with the help of Row Operations.

Row Operation 1: multiply the 1st row by -1/3

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 1 & 5 & -1 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

Row Operation 2: add -1 times the 1st row to the 2nd row

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

Row Operation 3: add 1 times the 1st row to the 3rd row

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ 0 & \frac{32}{3} & - \frac{10}{3} & - \frac{5}{3} \end{array} \right]$

Row Operation 4: multiply the 2nd row by 3/16

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]$

Row Operation 5: add -32/3 times the 2nd row to the 3rd row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]$

Row Operation 6: multiply the 3rd row by -1/15

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 7: add -5/4 times the 3rd row to the 2nd row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 8: add 8/3 times the 3rd row to the 1st row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 9: add 1/3 times the 2nd row to the 1st row

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

The reduced row echelon form of the augmented matrix is

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

which corresponds to the system

$\left\begin{array}{ccccc}x_1&&+\frac{9}{16} x_3&=&0&\\&1x_2&-\frac{5}{16}x_3&=&0&\\&&0&=&1&\end{array}\right$

Equation 3 cannot be solved, therefore, the system has no solution.

7 0

3 years ago