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Mumz [18]
2 years ago
6

If line b is perpendicular to line a, and line c is perpendicular to line a, what is the slope of line c?

Mathematics
2 answers:
slega [8]2 years ago
8 0

we know that

If line b is perpendicular to line a, and line c is perpendicular to line a,

then

line b and line c are parallel

and two lines parallel have the same slope

so

<u>Find the slope of the line b</u>

Let

A(-3,-2)\\B(2,3)

The formula to calculate the slope between two points is equal to

m=\frac{(y2-y1)}{(x2-x1)}

(x1,y1)=(-3,-2)\\(x2,y2)=(2,3)

substitute

m=\frac{(3+2)}{(2+3)}

m=\frac{(5)}{(5)}

m=1

therefore

<u>the answer is</u>

<u>the slope of the line c is</u>

mc=1

jeyben [28]2 years ago
5 0

Answer:

the answer is 1

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A financial advisor tells you that you can make your child a millionaire if you just start saving early. You decide to put an eq
Blababa [14]

Answer:

$12159 per year.

Step-by-step explanation:

If I invest $x each year at the simple interest of 7.5%, then the first $x will grow for 35 years, the second $x will grow for 34 years and so on.

So, the total amount that will grow after 35 years by investing $x at the start of each year at the rate of 7.5% simple interest will be given by

x( 1 + \frac{35 \times 7.5}{100}) + x( 1 + \frac{34 \times 7.5}{100}) + x( 1 + \frac{33 \times 7.5}{100}) + ......... + x( 1 + \frac{1 \times 7.5}{100})

= 35x + \frac{x \times 7.5}{100} [35 + 34 + 33 + ......... + 1]

= 35x + \frac{x \times 7.5}{100} [\frac{1}{2} (35) (35 + 1)]

{Since sum of n natural numbers is given by \frac{1}{2} (n)(n + 1)}

= 35x + 47.25x

= 82.25x

Now, given that the final amount will be i million dollars = $1000000

So, 82.25x = 1000000

⇒ x = $12,158. 05 ≈ $12159

Therefore. I have to invest $12159 per year. (Answer)

5 0
2 years ago
A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the
lbvjy [14]

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

Step-by-step explanation:

(a)

Let <em>X</em> = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

<em>X</em> = 610

<em>n</em> = 1000

Confidence level = 90%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{610}{1000}=0.61

The critical value of <em>z</em> for a 90% confidence level is:

z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

<u>Interpretation</u>:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let <em>X</em> = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

<em>X</em> = 440

<em>n</em> = 1000

Confidence level = 95%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{440}{1000}=0.44

The critical value of <em>z</em> for a 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend  is (0.42, 0.46).

<u>Interpretation</u>:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

  1. The confidence level
  2. Sample size
  3. Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of <em>z</em> in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

4 0
2 years ago
I need help ASAP!!!!!When bisecting an angle, three arcs are drawn. One is used to find points on the rays of the angle, and the
user100 [1]

Answer:

B

Step-by-step explanation:

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Sam and Edna have 56 marbles together Edna has 6 times more marbles than Sam. How many marbles does sam have ?
Elis [28]
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Read 2 more answers
The rabbit population on a small island is observed to be given by the function P(t) = 130t − 0.3t^4 + 1100 where t is the time
Montano1993 [528]
The maximum occurs when the derivative of the function is equal to zero.
P(t)=-0.3t^{4}+130t+1100 \\ P'(t)=-1.2t^{3}+130 \\ 0=-1.2t^{3}+130 \\ 1.2t^{3}=130 \\ t^{3}= \frac{325}{3}  \\ t=4.76702
Then evaluate the function for that time to find the maximum population.
P(t)=-0.3t^{4}+130t \\ P(4.76702)=-0.3*4.76702^{4}+130*4.76702+1100 \\ P(4.76702)=1564.79201
Depending on the teacher, the "correct" answer will either be the exact decimal answer or the greatest integer of that value since you cannot have part of a rabbit.
7 0
2 years ago
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