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3 years ago

I need answers asap will mark the first person to get it right brainlist!

Mathematics

2 answers:

Sergeeva-Olga [200]3 years ago

4 0

165 -25 =140
140/7= 20.
20 hours

serious [3.7K]3 years ago

3 0

Answer:

Step-by-step explanation:

what are the deopdowns

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The fifth graders collected 397.25 pounds of paper por the recycling drive. The sixth graders collected 431 pounds of paper. How

Snowcat [4.5K]

431 - 397.25 = 33.75 pounds of paper.

The sixth graders collected 33.75 more pounds of paper than the fifth graders.

4 0

3 years ago

What is the x intercept that is perpendicular to y=4/3x+1 and passes through (-4,9)

kolezko [41]

Well, we know the line is perpendicular to that one above.... what is the slope of that one anyway? well, notice, the equation is already in slope-intercept form $\bf y=\stackrel{slope}{\cfrac{4}{3}}x+1$ .

so, we're looking for the equation of a line perpendicular to that one, now, since that one has a slope of 4/3, a perpendicular line will have a negative reciprocal slope to that one,

$\bf \textit{perpendicular, negative-reciprocal slope for}\quad \cfrac{4}{3}\\\\
negative\implies -\cfrac{4}{ 3}\qquad reciprocal\implies - \cfrac{ 3}{4}$

so, what is the equation of a line whose slope is -3/4 and runs through -4,9?

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1\\
&&(~ -4 &,& 9~)
\end{array}
\\\\\\
% slope = m
slope = m\implies -\cfrac{3}{4}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-9=-\cfrac{3}{4}[x-(-4)]
\\\\\\
y-9=-\cfrac{3}{4}(x+4)$

now, the x-intercept for any function is found by zeroing out the "y" and solving for "x", thus

$\bf y-9=-\cfrac{3}{4}(x+4)\implies 0-9=-\cfrac{3}{4}(x+4)\implies -9=-\cfrac{3x}{4}-3
\\\\\\
-6=-\cfrac{3x}{4}\implies -24=-3x\implies \cfrac{-24}{-3}=x\implies 8=x$

x = 8, y = 0 ( 8 , 0 )

5 0

3 years ago

A sheet of paper has an area of 200 square inches. What dimensions will give the sheet the smallest perimeter?

PSYCHO15rus [73]

If it was 20 inches x 10 inches the perimeter would be 60 inches. which is the smallest possible perimeter with whole numbers

3 0

3 years ago

An education firm measures the high school dropout rate as the percentage of 16 to 24 year olds who are not enrolled in school a

posledela

Answer:

The value of the test statistic $z = -0.6606$

Step-by-step explanation:

From the question we are told that

The high dropout rate is $\mu = 6.1$ % $= 0.061$

The sample size is $n = 1000$

The number of dropouts $x = 56$

The probability of having a dropout in 1000 people $\= x = \frac{56}{1000} = 0.056$

Now setting up Test Hypothesis

Null $H_o : p = 0.061$

Alternative $H_a : p < 0.061$

The Test statistics is mathematically represented as

$z = \frac{\= x - p}{\sqrt{\frac{p(1 -p)}{n} } }$

substituting values

$z = \frac{0.056 - 0.061}{\sqrt{\frac{0.061(1 -0.061)}{1000} } }$

$z = -0.6606$

6 0

3 years ago

Solve the system of equations. −2x+5y =−35 7x+2y =25

Otrada [13]

Answer:

The equations have one solution at (5, -5).

Step-by-step explanation:

We are given a system of equations:

$\displaystyle{\left \{ {{-2x+5y=-35} \atop {7x+2y=25}} \right.}$

This system of equations can be solved in three different ways:

Graphing the equations (method used)
Substituting values into the equations
Eliminating variables from the equations

Graphing the Equations

We need to solve each equation and place it in slope-intercept form first. Slope-intercept form is $\text{y = mx + b}$ .

Equation 1 is $-2x+5y = -35$ . We need to isolate y.

$\displaystyle{-2x + 5y = -35}\\\\5y = 2x - 35\\\\\frac{5y}{5} = \frac{2x - 35}{5}\\\\y = \frac{2}{5}x - 7$

Equation 1 is now $y=\frac{2}{5}x-7$ .

Equation 2 also needs y to be isolated.

$\displaystyle{7x+2y=25}\\\\2y=-7x+25\\\\\frac{2y}{2}=\frac{-7x+25}{2}\\\\y = -\frac{7}{2}x + \frac{25}{2}$

Equation 2 is now $y=-\frac{7}{2}x+\frac{25}{2}$ .

Now, we can graph both of these using a data table and plotting points on the graph. If the two lines intersect at a point, this is a solution for the system of equations.

The table below has unsolved y-values - we need to insert the value of x and solve for y and input these values in the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & a \\ \cline{1-2} 1 & b \\ \cline{1-2} 2 & c \\ \cline{1-2} 3 & d \\ \cline{1-2} 4 & e \\ \cline{1-2} 5 & f \\ \cline{1-2} \end{array}$

$\bullet \ \text{For x = 0,}$

$\displaystyle{y = \frac{2}{5}(0) - 7}\\\\y = 0 - 7\\\\y = -7$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=\frac{2}{5}(1)-7}\\\\y=\frac{2}{5}-7\\\\y = -\frac{33}{5}$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=\frac{2}{5}(2)-7}\\\\y = \frac{4}{5}-7\\\\y = -\frac{31}{5}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=\frac{2}{5}(3)-7}\\\\y= \frac{6}{5}-7\\\\y=-\frac{29}{5}$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=\frac{2}{5}(4)-7}\\\\y = \frac{8}{5}-7\\\\y=-\frac{27}{5}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=\frac{2}{5}(5)-7}\\\\y=2-7\\\\y=-5$

Now, we can place these values in our table.

As we can see in our table, the rate of decrease is $-\frac{2}{5}$ . In case we need to determine more values, we can easily either replace x with a new value in the equation or just subtract $-\frac{2}{5}$ from the previous value.

For Equation 2, we need to use the same process. Equation 2 has been resolved to be $y=-\frac{7}{2}x+\frac{25}{2}$ . Therefore, we just use the same process as before to solve for the values.

$\bullet \ \text{For x = 0,}$

$\displaystyle{y=-\frac{7}{2}(0)+\frac{25}{2}}\\\\y = 0 + \frac{25}{2}\\\\y = \frac{25}{2}$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=-\frac{7}{2}(1)+\frac{25}{2}}\\\\y = -\frac{7}{2} + \frac{25}{2}\\\\y = 9$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=-\frac{7}{2}(2)+\frac{25}{2}}\\\\y = -7+\frac{25}{2}\\\\y = \frac{11}{2}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=-\frac{7}{2}(3)+\frac{25}{2}}\\\\y = -\frac{21}{2}+\frac{25}{2}\\\\y = 2$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=-\frac{7}{2}(4)+\frac{25}{2}}\\\\y=-14+\frac{25}{2}\\\\y = -\frac{3}{2}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=-\frac{7}{2}(5)+\frac{25}{2}}\\\\y = -\frac{35}{2}+\frac{25}{2}\\\\y = -5$

And now, we place these values into the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

When we compare our two tables, we can see that we have one similarity - the points are the same at x = 5.

Equation 1 Equation 2

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$ $\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

Therefore, using this data, we have one solution at (5, -5).

4 0

3 years ago