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djyliett [7]
3 years ago
11

Evan has graphed the point (5, 7) and he wants to reflect it over the line y=-2/5x+6. He predicts that the reflected point will

have coordinates (2, 2). Without graphing, can you confirm his answer or show that he cannot be correct?
Mathematics
1 answer:
Whitepunk [10]3 years ago
7 0

Given that Evan has graphed the point (5, 7) and he wants to reflect it over the line y=-2/5x+6. He predicts that the reflected point will have coordinates (2, 2).

Now we have to check if his answer is correct or not without graphing. Yes this is possible without graphing.

If his answer is correct then the line passing through points (5,7) and (2,2) will be perpendicular to the given line.

slope of line passing through the points (5,7) and (2,2) is given by :

m=\frac{y_2-y_1}{x_2-x_1}

m=\frac{7-2}{5-2}

m=\frac{5}{3}

slope of the given line y=-\frac{2}{5}x+6 is -\frac{2}{5}

We know that If the two lines are perpendicular then product of their slopes equals -1

-\frac{2}{5}*\frac{5}{3} = -\frac{2}{3}

Which is not -1. That means both lines are not perpendicular.

Hence his answer is wrong.

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The table below shows the scores of a group of students on a 10-point quiz.Test ScoreFrequency3 54 15 16 27 28 49 310 1The mean
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Explanation

Part A

We can use the formula below to find the mean.

\text{Mean}=\sum ^{}_{}(\frac{fx}{f})

We will then have;

\begin{gathered} \text{Mean = }\frac{5\times3+4\times1+5\times1+6\times2+7\times2+8\times4+9\times3+10\times1}{19}=\frac{15+4+5+12+14+32+27+10}{19} \\ =\frac{119}{19}=6.2632 \end{gathered}

Mean = 6.2632

Part B

Since the data set is odd.

\begin{gathered} \text{Median}=\frac{(n+1)}{2}^{th}\text{observation} \\ =\text{Value of }\frac{\text{(19+1)}}{2}^{th}\text{observation} \\ =\text{value of 10th observation} \end{gathered}

Answer: From the column of cumulative frequency cf, the median is 7

5 0
1 year ago
6.1.3
Nata [24]

Answer:

The requirements that are necessary for a normal probability distribution to be a standard normal probability distribution are <em>µ</em> = 0 and <em>σ</em> = 1.

Step-by-step explanation:

A normal-distribution is an accurate symmetric-distribution of experimental data-values.  

If we create a histogram on data-values that are normally distributed, the figure of columns form a symmetrical bell shape.  

If X \sim N (µ, σ²), then Z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z \sim N (0, 1).

The distribution of these z-variates is known as the standard normal distribution.

Thus, the requirements that are necessary for a normal probability distribution to be a standard normal probability distribution are <em>µ</em> = 0 and <em>σ</em> = 1.

8 0
3 years ago
Solve for x ~<br><img src="https://tex.z-dn.net/?f=6x%20-%2036%20%3D%2012" id="TexFormula1" title="6x - 36 = 12" alt="6x - 36 =
Shtirlitz [24]

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\large \tt \:  \:  \:  \:  \:  \: ⇝ 6x = 12 + 36

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\large \tt \:  \:  \:  \:  \:  \: ⇝ 6x = 48

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-_- Hope Helps!¡!¡!¡!¡!

7 0
2 years ago
Read 2 more answers
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