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ruslelena [56]
2 years ago
11

A construction company can remove 3/4 tons of dirt from a construction site each hour. how long will it take them to remove 36 t

ons of dirt from the site? (Answer needs to be in simplest form)
Mathematics
1 answer:
Lubov Fominskaja [6]2 years ago
3 0

Answer:

48hours

Step-by-step explanation:

36÷(3/4)

36*(4/3)

144/3

48

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In which direction must the graph of f(x) = |x| be shifted to produce the graph of g(x) = |x − 3|?
ElenaW [278]

Answer:

In which direction must the graph of f(x) = |x| be shifted to produce the graph of g(x) = |x − 3|?

A. left

B. down

C. up

<em><u>✓</u></em><em><u>D. right</u></em>

Right is the right answer.

6 0
3 years ago
Read 2 more answers
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
2 years ago
Chris wanted to know how likely he is to win at his favorite carnival game. He conducted 50 tests and won 15 times. What is the
aleksklad [387]

Answer:

b.) 0.30

Step-by-step explanation:

15/50 = 0.3

3 0
2 years ago
In the last quarter of​ 2007, a group of 64 mutual funds had a mean return of 2.2 2.2​% with a standard deviation of 6.3 6.3​%.
Svet_ta [14]

Answer:

a) 0.625; b) 16.879; c) 8.442

Step-by-step explanation:

Since this is a normal distribution, we want the value of the mean, μ:  2.2; and the value of the standard deviation, σ:  6.3.

For the 40th percentile, we look in a z chart.  We want to find the value as close to 0.40 as we can get; this is 0.4013, and it corresponds to a z score of z = -0.25.

Our formula for a z score is z=\frac{X-\mu}{\sigma}.  Using our values, we have

-0.25 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(-0.25) = X-2.2

-1.575 = X-2.2

Add 2.2 to each side:

-1.575+2.2 = X-2.2+2.2

0.625 = X

For the 99th percentile, the value in the z chart closest to 0.99 is 0.9901, which corresponds to a z score of z = 2.33:

2.33 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(2.33) = X-2.2

14.679 = X-2.2

Add 2.2 to each side:

14.679+2.2 = X-2.2+2.2

16.879 = X

For the IQR, we find the values for the 75th percentile (Q3) and the 25th percentile (Q1).  The value in a z chart closest to 0.75 is 0.7486, which corresponds to a z score of z = 0.67:

0.67 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(0.67) = X-2.2

4.221 = X-2.2

Add 2.2 to each side:

4.221+2.2 = X-2.2+2.2

6.421 = X

The value in a z chart closest to 0.25 is 0.2514, which corresponds to a z score of z = -0.67:

-0.67 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(-0.67) = X-2.2

-4.221 = X-2.2

Add 2.2 to each side:

-4.221+2.2 = X-2.2+2.2

-2.021 = X

This makes the interquartile range

6.421--2.021 = 8.442

8 0
3 years ago
Can someone help me with this work please
tangare [24]

Answer:

-See below

Step-by.step explanation:

In the first column, on the left of the vertical line, you place the first digit of the number , then on the second row on you place the second digit.

So on the second row you have the entries:

1 | 2 2 4 representing  12, 12 and 14.

On the first  row the 2 0ne digit numbers 3 and 8 are represented by

0 | 3 8.

Similarly the last 2 rows are:

2 |  0 1 3 6

3 |  4

8 0
2 years ago
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