All categories

2 years ago

The verticles of a triangle are the points R(3,c),Q(9,2) and R(3c,11) where c is constant. Given that angle PQR is 90

Mathematics

1 answer:

IceJOKER [234]2 years ago

5 0

Answer:

Step-by-step explanation:

Find c if ∠PQR = 90°?

I will ASSUME you mean point P is at (3. c)

slope of PQ is (2 - c) / (9 - 3) = (2 - c) / 6

slope of QR is (11 - 2) / (3c - 9) = 9 / (3c - 9)

perpendicular lines have negative reciprocal slopes.

(2 - c) / 6 = -1(3c - 9)/9

9(2 - c) = -6(3c - 9)

18 - 9c = -18c + 54

9c = 36

c = 4

You might be interested in

PLEASE HELP IM SO STUCK 20 POINTS

prohojiy [21]

Answer: X = 2

Step-by-step explanation: The correct answer to this question is that x=2. Regardless of what point you go to on the line, it always has a value of 2 because it is going vertically (up and down) at 2.

5 0

3 years ago

Please HELP ME!!!! This is really confusing. HOW DO I DO THIS?!?!?!

Svetradugi [14.3K]

Answer:

Flying Disc should be 72

Step-by-step explanation:

circumference = πd

HD Coin C = 3π

9.42477796077

Flying Disc. C = 23π
72.2566310326

Jar Lid C = 8π

25.1327412287

Flower Pot C = 15π

47.1238898038

8 0

1 year ago

Kelko will rent a car for the weekend. She can choose one of two plans. The first plan has an initial fee of $57 and cost an add

Delicious77 [7]

Questions (contd)

(a) For what amount of driving do the two plans cost the same?

(b) What is the cost when the two plans cost the same?

Answer:

(a) 100 miles

(b) $65

Step-by-step explanation:

Given

Plan 1:

$Initial\ Fee = \$57$

$Additional = \$0.08$ per mile

Plan 2:

$Initial\ Fee = \$50$

$Additional = \$0.10$ per mile

Solving (a): Number of miles when both plans are equal

Represent the distance with x and the cost with y

So:

Plan 1:

$y = 57 + 0.08 * x$

Plan 2:

$y = 50 + 0.10 * x$

To solve (a), we equate both plans together; i.e.

$y = y$

$57 + 0.08x = 50 + 0.10x$

Collect Like Terms

$0.08x - 0.10x= 50 - 57$

$-0.02x= -2$

Solve for x

$x = -2/-0.02$

$x = 100$

Hence, 100 mile would cost both plans the same

Solving (b): Cost when both plans are the same:

In this case, we simply substitute 100 for x in any of the y equation.

$y = 57 + 0.08 * x$

$y = 57 + 0.08 * 100$

$y = 57 + 8$

$y = 65$

Hence, the amount is $65

7 0

2 years ago

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then

$\\ P(z$

However, we are looking for P(z>1.56), which is the complement probability of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

5 0

2 years ago

Maya has 120 caramel apples to sell. Each caramel apple is covered with one toppin.

EleoNora [17]

they got sold by her

Step-by-step explanation:

lol she sold her apples to people

5 0

2 years ago

The verticles of a triangle are the points R(3,c),Q(9,2) and R(3c,11) where c is constant. Given that angle PQR is 90​

The verticles of a triangle are the points R(3,c),Q(9,2) and R(3c,11) where c is constant. Given that angle PQR is 90