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2 years ago

It takes 41 minutes for 9 people to paint 9 walls.

Mathematics

1 answer:

lord [1]2 years ago

7 0

Answer:

it take 41 minutes to 14 people to paint 14 walls

Step-by-step explanation:

we know that

It takes 41 minutes for 9 people to paint 9 walls.

Divide the number of people by the number of walls

$9/9=1\frac{people}{wall}$

it take 41 minutes to one people to paint one wall

therefore

it take 41 minutes to 14 people to paint 14 walls

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This is Matrix for pre calc

gavmur [86]

The given system of equations in augmented matrix form is

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]$

If you need to solve this, first get the matrix in RREF:

Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]$

Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]$

Add 164(row 3) to -91(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]$

Multiply row 4 by 1/13080:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]$

Add -153(row 4) to row 3:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]$

Multiply row 3 by -1/91:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add 6(row 3) and -8(row 4) to row 2:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 2 by 1/5:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

$\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 1 by 1/3:

$\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

So the solution to this system is $(w,x,y,z)=(1,-2,4,-3)$ .

6 0

3 years ago

0.99 0.89 7/8 least to greatest

katovenus [111]

From least to greatest 7/8, 0.89, 0.99

6 0

3 years ago

Hailey wants to hire someone to paint her house. She researched several independent painters who provided information about the

Reika [66]

Answer:

my name is hailey

Step-by-step explanation:

6 0

2 years ago

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f

Sloan [31]

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen who carry a credit card balance.

The formula for the interval is:

p'₁± $Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }$

$Z_{1-\alpha /2}= Z_{0.95}= 1.648$

0.37±1.648* $\sqrt{\frac{0.37*0.63}{1000} }$

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂± $Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }$

0.48±1.648* $\sqrt{\frac{0.48*0.52}{1000} }$

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: $\sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015$

Seniors: $\sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016$

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

8 0

3 years ago

Help pleas Describe the power function f(x)=x^5. What does it look like? Where does it cross the x- and y-axis? How many turns d

dezoksy [38]

Answer:53=125

Step-by-step explanation:For an increasing function like this, the end behavior at the right "end" is going to infinity. Written like: as x→∞,y→∞ . That means that large powers of 5 will continue to grow larger and head toward infinity. For example, 53=125 .

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2 years ago