The sum of 1/6, 2/3, and 1/4 is

Towns P,Q,R and S are shown. Q is 35 km due East of P S is 15km due West of P R is 15km due South of P Work out the bearing of R

Anni [7]

Answer:

Part A

The bearing of the point 'R' from 'S' is 225°

Part B

The bearing from R to Q is approximately 293.2°

Step-by-step explanation:

The location of the point 'Q' = 35 km due East of P

The location of the point 'S' = 15 km due West of P

The location of the 'R' = 15 km due south of 'P'

Part A

To work out the distance from 'R' to 'S', we note that the points 'R', 'S', and 'P' form a right triangle, therefore, given that the legs RP and SP are at right angles (point 'S' is due west and point 'R' is due south), we have that the side RS is the hypotenuse side and ∠RPS = 90° and given that $\overline{RP}$ = $\overline{SP}$ , the right triangle ΔRPS is an isosceles right triangle

∴ ∠PRS = ∠PSR = 45°

The bearing of the point 'R' from 'S' measured from the north of 'R' = 180° + 45° = 225°

Part B

∠PRQ = arctan(35/15) ≈ 66.8°

Therefore the bearing from R to Q = 270 + 90 - 66.8 ≈ 293.2°

6 0

3 years ago

Guys I just went poop and there's no toilet paper please help what do I do oh my godddd

iogann1982 [59]

Answer:

Use water to clean yourself and than just wash your hands afterward lol

Step-by-step explanation:

8 0

3 years ago

3) An open top box is to be constructed out of a 90 inch by 70 inch piece of cardboard by cutting squares out of the corners and

Elena L [17]

Answer:

a) The volume in cubic inches is 36000

b) The volume in cubic feet is 125/6

c) The volume in cubic yard is 125/162

Step-by-step explanation:

* Lets study the information of the problem to solve it

- The dimensions of the piece of cardboard are 90 inches by 70 inches

- The side of the cutting square is 15 inches

- The squares are cutting from each corner

∴ Each dimension of the cardboard will decrease by 2 × 15 inches

∴ The new dimensions of the piece of cardboard are;

90 - (15 × 2) = 90 - 30 = 60 inches

70 - (2 × 15) = 70 - 30 = 40 inches

- The dimensions of the box will be:

# Length = 60 inches

# width = 40 inches

# height = 15 inches

- The volume of any box with three different dimensions is

V = Length × width × height

∵ The length = 60 inches

∵ The width = 40 inches

∵ The height = 15 inches

∴ V = 60 × 40 × 15 = 36000 inches³

a) The volume in cubic inches is 36000

* Now lets revise how to change from inch to feet

- 1 foot = 12 inches

∵ 1 foot = 12 inches

∴ 1 foot³ = (12)³ inches³

∴ 1 foot³ = 1728 inches³

∵ The volume of the box is 36000 inches³

∴ The volume of the box in cubic feet = 36000 ÷ 1728 = 125/6

b) The volume in cubic feet is 125/6

* Now lets revise how to change from feet to yard

- 1 yard = 3 feet

∵ 1 yard = 3 feet

∴ 1 yard³ = (3)³ feet³

∴ 1 yard³ = 27 feet³

∵ The volume of the box is 125/6 feet³

∴ The volume of the box in cubic yard = 125/6 ÷ 27 = 125/162

c) The volume in cubic yard is 125/162

8 0

3 years ago

Read 2 more answers

The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen

vfiekz [6]

Answer:

a) 0.2581

b) 0.4148

c) 17

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$p = 0.75$

a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X = 9)$ when $n = 12$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{12,9}.(0.75)^{9}.(0.25)^{3} = 0.2581$

b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X \geq 16)$ when $n = 20$