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Brilliant_brown [7]
3 years ago
10

Convert y=-1/4x+4 in standard form

Mathematics
1 answer:
kondor19780726 [428]3 years ago
3 0
Idk but I think it is 3 so hope it is right so please accept
You might be interested in
Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

7 0
3 years ago
Please help!!!!! 10 points!!
IrinaK [193]

Answer:

Both calculated the slope as 1/3

Step-by-step explanation:

Avis:


Rise = 4

Run = 12

Rise goes over the run:   4/12

4/12 simplifies into 1/3


Alec:


Rise= 2

Run= 6

Rise goes over the run:   2/6

2/6 simplifies into 1/3

4 0
3 years ago
Read 2 more answers
Find the next three sequence of 1,3, 9, 27,<br>​
Andreyy89

Answer: 81, 243, 729

Step-by-step explanation:

<u>STEP ONE: Find a pattern in the sequence</u>

3 ÷ 1 = 3

9 ÷ 3 = 3

27 ÷ 9 = 3

As we can see from the given sequence, each number multiplies 3 to get the next term.

<u>STEP TWO: Find the next three terms</u>

27 × 3 = 81

81 × 3 = 243

243 × 3 = 729

Hope this helps!! :)

Please let me know if you have any questions

4 0
3 years ago
Pls help! Will give brainliest
mariarad [96]
First is -1 and 4, second should be no solution and the third is 2 and 6
5 0
3 years ago
The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge
Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

5x+4y+5z=-1

x+y+2z=1

2x+y-z=-3

Firs we find determinant of system of equations

Let a matrix A=\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right] and B=\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]

\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}

\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply R_1\rightarrow R_1-4R_2 and R_3\rightarrow R_3-2R_2

\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]:\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]

ApplyR_2\rightarrow R_2-R_1

\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]:\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]

Apply R_3\rightarrow R_3+R_2

\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]:\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]

Apply R_3\rightarrow- \frac{1}{2} and R_2\rightarrow R_2-R_3

\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]:\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]

Apply R_1\rightarrow R_1-R_3

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]:\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

4 0
3 years ago
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