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3 years ago

A rectangle has one side of llcm and

Mathematics

1 answer:

Marizza181 [45]3 years ago

8 0

The opposite side of the 11cm will also be 11cm and the 2 others will both be 8 cm.

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Express (13/15 - 7/10) as a percentage

sasho [114]

Answer:

(13/15 - 7/10)

take lcm

<u>13*2 - 7*3</u> = 5/30*100= 50/3%

Step-by-step explanation:

8 0

2 years ago

Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round y

tino4ka555 [31]

Answer:

0.8762 or 87.62%

Step-by-step explanation:

Since our mean is μ=14.3 and our standard deviation is σ=3.7. If we're trying to figure out what percentage is P(10 ≤ x ≤ 26) equal to we must first calculate our z values as such:

$z=\frac{x-\mu}{\sigma}$

Our x value ranges from 10 to 26 therefore let x=10 and we obtain:

$z=\frac{10-14.3}{3.7} =-1.16\\$

If we look at our z-table we find that the probability associated with a z value of -1.16 is 0.1230 meaning 12.30%.

Now let's calculate the z value when x = 26 and so:

$z=\frac{26-14.3}{3.7}=3.16\\$

Similarly, we use the z-table again and find that the probability associated with a z value of 3.16 is 0.9992 meaning 99.92%.

Now we want to find the probability in between 10 and 26 so we will now subtract the upper limit minus the lower limit in P(10 ≤ x ≤ 26) therefore:

0.9992 - 0.1230 = 0.8762

or 87.62%

7 0

3 years ago

Solve the system of equations.<br><br><br><br> −2x+5y =−35<br> 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

<u>Graphing the Equations</u>

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

Each month, Liz pays $35 to her phone company just to use the phone. Each text she sends costs her an additional $0.05. In March

Firdavs [7]

<h3><u>Answer:</u></h3><h3><u>752 & 617</u></h3><h3 /><h3><u>Explanation:</u></h3><h3><u>So if Liz pays $35 each month just to use the phone, we could subtract 35 from the total bill that month to get the the total cost she spent on text messages.</u></h3><h3 /><h3><u>March:$72.60-$35=$37.60</u></h3><h3><u>April: $65.85-$35=$30.85</u></h3><h3 /><h3><u>We can see that in March Liz spent $37.60 on texts in total and in April she spent $30.85 on texts in total. All we have to do is divide the amount of money she spent on texts( $37.60 & $30.85) by the cost of one text message($0.05) to get the amount of texts she sent that month.</u></h3><h3 /><h3><u>March: $37.60/$0.05=752</u></h3><h3><u>April: $30.85/$0.05=617</u></h3><h3 /><h3><u>Therefore, she sent 752 texts in March and 617 texts in April.</u></h3><h3 /><h3><u>~FrxziteTheLxoser~ I hoped I help you :)</u></h3>

<u>If i helped you please give me brainIIlest!</u>

3 0

3 years ago

Describe lengths of three segments that could not be used to form a triangle. Segments with lengths of 5 in., 5 in., and ______

Vesna [10]

Answer:

$c = 11$