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1 year ago

Given that line s is perpendicular to line t, which statements must be true of the two lines? Check all that apply.

2 answers:

5 0

The basis to respond this question are:

1) Perpedicular lines form a 90° angle between them.

2) The product of the slopes of two any perpendicular lines is - 1.

So, from that basic knowledge you can analyze each option:

<span>a.Lines s and t have slopes that are opposite reciprocals.

TRUE. Tha comes the number 2 basic condition for the perpendicular lines.

slope_1 * slope_2 = - 1 => slope_1 = - 1 / slope_2, which is what opposite reciprocals means.

b.Lines s and t have the same slope.

FALSE. We have already stated the the slopes are opposite reciprocals.

c.The product of the slopes of s and t is equal to -1

TRUE: that is one of the basic statements that you need to know and handle.

d.The lines have the same steepness.

FALSE: the slope is a measure of steepness, so they have different steepness.

e.The lines have different y intercepts.

FALSE: the y intercepts may be equal or different. For example y = x + 2 and y = -x + 2 are perpendicular and both have the same y intercept, 2.

f.The lines never intersect.

FALSE: perpendicular lines always intersept (in a 90° angle).

g.The intersection of s and t forms right angle.

TRUE: right angle = 90°.

h.If the slope of s is 6, the slope of t is -6

FALSE. - 6 is not the opposite reciprocal of 6. The opposite reciprocal of 6 is - 1/6.

So, the right choices are a, c and g.
</span>

Reil [10]1 year ago

3 0

Answer:

A C G

Step-by-step explanation:

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Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

$P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}$

And if we want to find the joint probability we just need to do this:

$P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)$

And if we multiply the probabilities founded we got:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

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1 year ago

The graph below represents the possible dimensions of a horse corral to be constructed behind a barn. The width of the corral is

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I think its C

Step-by-step explanation:

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In a recent survey, the proportion of adults who indicated mystery as their favorite type of book was 0.325. Two simulations wil

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Answer:

C) The centers will roughly be equal, and the variability of simulation A will be less than the variability of simulation B.

Step-by-step explanation:

The center for Simulation A and Simulation B will be roughly equal.

Overall Sample size of Simulation A = 1500 * 100 = 150000

Overall Sample size of Simulation B = 2000 * 50 = 100000

Since the sample size for Simulation A is greater, the variability of Simulation will be less.

Therefore, The answer is C) The centers will roughly be equal, and the variability of simulation A will be less than the variability of simulation B.

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2 years ago

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

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Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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