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3 years ago

Pentagon ABCDE and pentagon A'B'C'D'E' are shown on the coordinate plane below:

Mathematics

1 answer:

s2008m [1.1K]3 years ago

7 0

Answer:

<h2>The third option is correct.</h2>

Step-by-step explanation:

The line D'E' is parallel to the y-axis.

Hence, it is clear that the image is reflected across the y-axis.

If you see closely the given image, you can see that the y ordinates of each point of the Pentagon has shifted 8 points towards the positive direction of y-axis.

Hence, the transformation for the y-coordinates will be Y = y + 8.

Now if you check the transformation for the x-coordinates, you will observe that the coordinates of D and E will be lie on x = -1. After reflection, they will lie on x = 1.

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What is 5.316 - 1.942 btw (show ur work) :)

erastovalidia [21]

Answer:

3.374

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4\\$

$\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}\\$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&11&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&\linethrough{1}&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{3}&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}$

=3.374

6 0

2 years ago

Duck #1 lays eggs whose weights are normally distributed with a mean of 70gramsand a standard deviation of 6 grams.Duck #2 lays

Genrish500 [490]

Answer:

26.11% probability that Duck #2’s egg weighs more than Duck #1’s egg

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If an egg is randomly chosen from each duck, what is the probability that Duck #2’s egg weighs more than Duck #1’s egg?

Duck #2 egg will weigh more if the subtraction of duck's 2 egg by duck's 1 egg is larger than 0.

When we subtract normal distributions, the mean is the subtraction of the means. So

$\mu = 65 - 70 = -5$

The standard deviation is the square root of the sum of the variances. So

$\sigma = \sqrt{6^2+5^2} = \sqrt{61} = 7.81$

Now, we have to find 1 subtracted by the pvalue of Z when X = 0. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0 - (-5)}{7.81}$

$Z = 0.64$

$Z = 0.64$ has a pvalue of 0.7389

1 - 0.7389 = 0.2611

26.11% probability that Duck #2’s egg weighs more than Duck #1’s egg

4 0

3 years ago

If f(x) = 2x^2 + 1 and g(x) = x^2 - 7, find (f+g)(x)

jarptica [38.1K]

(f + g)(x) means f(x) + g(x)

So, you just input the function of f(x) and g(x) altogether, then simplify it.
(f + g)(x)
= f(x) + g(x)
= 2x² + 1 + x² - 7
= 2x² + x² + 1 - 7
= 3x² - 6

The answer is b

5 0

3 years ago

Because of her past convictions for mail fraud and forgery, Jody has a 30% chance each year of having her tax returns audited. W

Flura [38]

Answer:

The probability that she will escape detection for at least three years is P=0.343.

Step-by-step explanation:

If Jody has a 30% chance each year of having her tax returns audited, she also has 70% chance each year of escaping detection.

The probability of this happening 3 years in a row is:

$P_n=q^n=0.7^3=0.343$