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2 years ago

What percentage is 650 m to 4 km.

Mathematics

1 answer:

Nina [5.8K]2 years ago

3 0

Answer: 16.25%

Step-by-step explanation:

Change
650m = 0.65 km

Convert into percentage 0.65/ 4 x 100 = 16.25

You might be interested in

What is the solution to this system of equations y=x-2 y=-0.5x+4

gladu [14]

Answer:

Substitute y=x-2y=x−2 into y=-0.5x+4y=−0.5x+4.

x-2=-0.5x+4x−2=−0.5x+4

2 Solve for xx in x-2=-0.5x+4x−2=−0.5x+4.

x=4x=4

3 Substitute x=4x=4 into y=x-2y=x−2.

y=2y=2

4 Therefore,

\begin{aligned}&x=4\\&y=2\end{aligned}

x=4

y=2

Step-by-step explanation:

6 0

2 years ago

In some gymnastics meets, the score given to a gymnast is the mean of the judgesâ€™ scores after the highest and lowest scores h

Allisa [31]

Using the median concept, the correct statement is given as follows:

Her median score does not change.

<h3>What is the median of a data-set?</h3>

The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile, also called the middle value the data-set, not being affected by outliers.

For this problem, the data-set is given as follows:

7.50, 7.50, 7.75, 7.75, 8.00, 8.00, 8.00, 10.00.

There are 8 scores, hence the median is the <u>mean of the 4th and the 5th</u> scores, as follows:

(7.75 + 8)/2 = 7.875.

Removing the extreme values, the data-set will be given by:

7.50, 7.75, 7.75, 8.00, 8.00, 8.00.

Then it has 6 scores, hence the median is the <u>mean of the 3th and the 4th</u> scores, as follows:

(7.75 + 8)/2 = 7.875.

Same median, hence the correct option is:

Her median score does not change.

More can be learned about the median of a data-set at brainly.com/question/23923146

#SPJ1

7 0

2 years ago

Candice is taking a group nature hike in a park. There is a

slavikrds [6]

Answer:

500 miles

Step-by-step explanation:

81-11 is 70. 70 divided by .14 is 500 so that's 500 miles she can bike.

8 0

2 years ago

Read 2 more answers

Solve the system of equations by row-reduction. At each step, show clearly the symbol of the linear combinations that allow you

adell [148]

Answer:

1) The solution of the system is

$\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) The solution of the system is

$\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$

Step-by-step explanation:

1) To solve the system of equations

$\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right$

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

$\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 2: Swap rows 1 and 2

$\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 3: $\left(R_1=\frac{R_1}{6}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 5: $\left(R_2=\frac{R_2}{3}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 6: $\left(R_3=R_3+R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right]$

Step 7: $\left(R_3=\left(\frac{6}{37}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 8: $\left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 9: $\left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 10: Rewrite the system using the row reduced matrix:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) To solve the system of equations

$\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right$

using the row reduction method you must:

Step 1:

$\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 2: $\left(R_1=\frac{R_1}{4}\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 3: $\left(R_2=R_2-\left(2\right)R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 5: $\left(R_2=\left(2\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 6: $\left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 7: $\left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right]$

Step 8: $\left(R_3=\left(- \frac{2}{117}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 9: $\left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 10: $\left(R_2=R_2-\left(15\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 11:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$

8 0

3 years ago

The national mean sales price for a new one-family home is $181,900. A sample of 40 one-family homes in the south showed a sampl

SIZIF [17.4K]

Answer:

The null hypothesis is $\mu = \$181,900$

The alternative hypothesis is $\mu < \$ 181.900$

$t = -2.92$

$p-value = 0.0016948$

There no sufficient evidence to support the conclusion that the population mean sales prices for new one-family homes in the South is less expensive than the national mean of $181,900

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = \$ 181, 900$

The sample size is $n = 40$

The sample mean is $\= x = \$ 166,400$

The sample standard deviation is $s= \$ 33, 500$

The null hypothesis is $\mu = \$181,900$

The alternative hypothesis is $\mu < \$ 181.900$

Generally the test statistics is mathematically represented as

$t = \frac{ \= x - \mu }{ \frac{s}{\sqrt{n} } }$

=> $t = \frac{ 166400 - 181900 }{ \frac{33500}{\sqrt{40} } }$

=> $t = -2.92$

Generally the p-value is obtain from the z-table the value is

$p-value = P(Z < t ) = P(Z < -2.93) = 0.0016948$

=> $p-value = 0.0016948$

From the calculation we see that

$p-value > \alpha$ hence we fail to reject the null hypothesis

Thus there no sufficient evidence to support the conclusion that the population mean sales prices for new one-family homes in the South is less expensive than the national mean of $181,900

3 0

3 years ago