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3 years ago

Graph the image of this quadrilateral after a dilation with a scale factor of 4 centered at the origin

Mathematics

1 answer:

Brut [27]3 years ago

8 0

Answer: The image is attached.

Step-by-step explanation:

A Dilation is defined as a transformation in which the image and the pre-image have the same shape, but their sizes are different.

When the scale factor is greater than 1, the image obtained after the dilation is greater than the pre-image and it is an "Enlargement".

When the scale factor is is between 0 and 1, the image obtained after the dilation is smaller than the pre-image and it is an "Reduction".

In this case we know that the scale factor is:

$scale\ factor=4$

Since:

$4>1$

It is an Enlargement.

You can identify that the vertices of the pre-image shown in the figure, are:

$(-2,1)\\\\(-1,2)\\\\(0,0)\\\\(-1,-2)$

So you need to multiply the coordinates of those vertices by 4 in order to get the coordinates of the image;

$((-2)(4),1(4))=(-8,4)\\\\((-1)(4),(2)(4))=(-4,8)\\\\((0)(4),(0)(4)=(0,0)\\\\((-1)(4),(-2)(4))=(-4,-8)$

Now you can draw it (The image is attached).

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4x-3y=2 3x=5y=16 solve using substitution method

vodomira [7]

$\left\{\begin{array}{ccc}4x-3y=2&|\text{add 3y to both sides}\\3x+5y=16\end{array}\right\\\left\{\begin{array}{ccc}4x=3y+2&|\text{divide both sides by 4}\\3x+5y=16\end{array}\right\\\left\{\begin{array}{ccc}x=0.75y+0.5&(*)\\3x+5y=16\end{array}\right\\\\\text{substitute}\ (*)\ \text{to the second equation}\\\\3(0.75y+0.5)+5y=16\qquad\text{use distributive property}\\\\(3)(0.75x)+(3)(0.5)+5y=16\\\\2.25y+1.5+5y=16\qquad\text{subtract 1.5 from both sides}\\\\7.25y=14.5\qquad\text{divide both sides by 7.25}$

$\boxed{y=2}\\\\\text{substitute the value of y to}\ (*):\\\\x=0.75(2)+0.5\\\\x=1.5+0.5\\\\\boxed{x=2}\\\\Answer:\ x=2\ and\ y=2\to(2,\ 2)$

7 0

2 years ago

A College Alcohol Study has interviewed random samples of students at four-year colleges. In the most recent study, 494 of 1000

Vinil7 [7]

Answer:

The 95% confidence interval for the difference between the proportion of women who drink alcohol and the proportion of men who drink alcohol is (-0.102, -0.014) or (-10.2%, -1.4%).

Step-by-step explanation:

We want to calculate the bounds of a 95% confidence interval of the difference between proportions.

For a 95% CI, the critical value for z is z=1.96.

The sample 1 (women), of size n1=1000 has a proportion of p1=0.494.

$p_1=X_1/n_1=494/1000=0.494$

The sample 2 (men), of size n2=1000 has a proportion of p2=0.552.

$p_2=X_2/n_2=552/1000=0.552$

The difference between proportions is (p1-p2)=-0.058.

$p_d=p_1-p_2=0.494-0.552=-0.058$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{494+552}{1000+1000}=\dfrac{1046}{2000}=0.523$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.523*0.477}{1000}+\dfrac{0.523*0.477}{1000}}\\\\\\s_{p1-p2}=\sqrt{0.000249+0.000249}=\sqrt{0.000499}=0.022$