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Challenge fr y'all. solve this to get 50 points and brainliest coming up if you get the right answer

TEA [102]

Answer:

See below

Step-by-step explanation:

a)

i)

Use the law of cosines:

$AB=\sqrt{PA^2+PB^2-2*PA*PB*cosP}$
$AB=\sqrt{60^2+100^2-2*60*100*cos80} =\sqrt{11516.22} =107.31$

ii)

Use the law of sines:

sin ∠A / PB = sin ∠P / AB
sin ∠A / 100 = sin 80 / 107.31
sin ∠A = 100 sin 80 deg / 107.31
sin ∠A = 0.92
∠A = arcsin 0.92
∠A = 67° (rounded)

iii)

Bearing of B from A:

90° + (67° - 30°) =
127°

b)

PB is longer distance. Time to reach B:

t = d/s
t = 100/20 = 5 hours

The speed s of the slower ship:

s = d/t
s = 60/5 = 12 km/h

3 0

2 years ago

Read 2 more answers

HELP MEEEEEEEEEE SOOOOOO EASY PLEASE AND BRANLIEST TOOOOOO (07.04 LC)What is the solution to the equation fraction 1 over 3x = 6

nasty-shy [4]

Answer:

x = 2

divide each term by 3 and simplify.

7 0

3 years ago

Read 2 more answers

The mean number of hours of flying time for pilots at Continental Airlines is 49 hours per month. Assume that this mean was base

Effectus [21]

Answer:

a) 17.09 hours

b) The 95% confidence interval estimate of the population mean flying time for the Pilots is between 31.91 hours and 66.09 hours

Step-by-step explanation:

We have the standard deviation of the sample, so we use the t distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 49 - 1 = 48

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 48 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.0106

The margin of error is:

M = T*s = 2.0106*8.5 = 17.09

s is the standard deviation of the sample. 17.09 hours is the answer for a.

The lower end of the interval is the sample mean subtracted by M. So it is 49 - 17.09 = 31.91 hours

The upper end of the interval is the sample mean added to M. So it is 49 + 17.09 = 66.09 hours

The 95% confidence interval estimate of the population mean flying time for the Pilots is between 31.91 hours and 66.09 hours

8 0

3 years ago

Fred flips a coin, spins the spinner, and rolls a standard number cube. Find the probability that the coin will show tails, the

Inessa05 [86]

Answer:

The probability that the coin will show tails, the spinner will land on yellow or green (out of yellow, green, blue), and the cube will show a three is $\frac{4}{3}$

Step-by-step explanation:

Given : Fred flips a coin, spins the spinner, and rolls a standard number cube.

To find : The probability that the coin will show tails, the spinner will land on yellow or green (out of yellow, green, blue), and the cube will show a three ?

Solution :

First we find probability one by one,

1) Flips a coin - Head, Tail

Total number of outcome = 2

Favorable outcome (getting a tail) = 1

Probability that the coin will show tails is $P(T)=\frac{1}{2}$

2) Spins the spinner - yellow, green, blue

Total number of outcome = 3

Favorable outcome (land on yellow or green) = 2

Probability that the spinner will land on yellow or green $P(S)=\frac{2}{3}$

3) Rolls a standard number cube - 1,2,3,4,5,6

Total number of outcome = 6

Favorable outcome (show a three) = 1

Probability that the cube will show a three $P(C)=\frac{1}{6}$

The probability that the coin will show tails, the spinner will land on yellow or green (out of yellow, green, blue), and the cube will show a three is

$P=P(T)+P(S)+P(C)$

$P=\frac{1}{2}+\frac{2}{3}+\frac{1}{6}$

$P=\frac{3+4+1}{6}$

$P=\frac{8}{6}$

$P=\frac{4}{3}$

Therefore, The probability that the coin will show tails, the spinner will land on yellow or green (out of yellow, green, blue), and the cube will show a three is $\frac{4}{3}$

5 0

3 years ago

Find x please!! a.10 b.32 c.116 d.16

AysviL [449]

9514 1404 393

Answer:

(d) 16

Step-by-step explanation:

Angles opposite sides of the same measure are congruent. Here, the triangle is isosceles, so the base angles are congruent:

2x = 32

x = 16 . . . . . . divide by 2°

5 0

3 years ago