Round 7905.68466806 to the nearest thousand.

Please... Solve this: the bearing of a tree from a house is 295°. Find the bearing of the house from the tree.

jonny [76]

Given parameters:

Bearing of the tree from the house = 295°

Unknown:

Bearing of the house from the tree = ?

Solution:

This is whole circle bearing problem(wcb). There are different ways of representing directions from one place to another. In whole circle bearing, the value of the bearing varies from 0° to 360° in the clockwise direction.

To find the back bearing in this regard, simple deduct the forward bearing from 360°;

Backward bearing = 360° -295°

= 65°

The bearing from house to the tree is 65°

4 0

2 years ago

A bookstore is hosts the stars from the Harry Potter

shepuryov [24]

Answer:

5985 can fit in the room!

Step-by-step explanation:

1) 150 * 100 = 15000

2) 6 * 6 = 36

3) 15000 - 36 = 14964

4) 14964 / 2.5 = 5985.6

5) But you can't fit almost half a person inside or create another whole person soo you round down to 5985

3 0

2 years ago

Read 2 more answers

The table shows data gathered by an environmental agency about the decreasing depth of a lake during the first few months of a d

EastWind [94]

(0,346)(2,344.8)
slope = (344.8 - 346) / (2 - 0) = -1.2 / 2 = -0.6

y = mx + b
slope(m) = -0.6
use either of ur points (0,346)...x = 0 and y = 346
now we sub and find b, the y int
346 = -0.6(0) + b
346 = b

so ur equation is : y = -0.6x + 346

after 4 weeks....x = 4
y = -0.6(4) + 346
y = -2.4 + 346
y = 343.6 <=== after 4 weeks it will be 343.6

5 0

2 years ago

Express x in term of y:x/7+2y=6

lions [1.4K]

Answer:

You have to multiply the denorminator to both sides in order to make x the subject :

$\frac{x}{7 + 2y} = 6$

$x = 6(7 + 2y)$

8 0

2 years ago

Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in

ASHA 777 [7]

Answer:

a) There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b) There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c) The expected number of Chrome users is 4.074.

d) The variance for the number of Chrome users is 3.2441.

The standard deviation for the number of Chrome users is 1.8011.

Step-by-step explanation:

For each Internet browser user, there are only two possible outcomes. Either they use Chrome, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

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 combinatios 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:

Google Chrome has a 20.37% share of the browser market. This means that $p = 0.2037$

20 Internet users are sampled, so $n = 20$ .

a.Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

This is P(X = 8).

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

$P(X = 8) = C_{20,8}.(0.2037)^{8}.(0.7963)^{12} = 0.0243$

There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b.Compute the probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

Either there are less than 3 Chrome users, or there are three or more. The sum of the probabilities of these events is decimal 1. So:

$P(X < 3) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{20,0}.(0.2037)^{0}.(0.7963)^{20} = 0.0105$

$P(X = 1) = C_{20,1}.(0.2037)^{1}.(0.7963)^{19} = 0.0538$

$P(X = 2) = C_{20,2}.(0.2037)^{2}.(0.7963)^{18} = 0.1307$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0105 + 0.0538 + 0.1307 = 0.1950$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.1950 = 0.8050$

There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c.For the sample of 20 Internet browser users, compute the expected number of Chrome users

We have that, for a binomial experiment:

$E(X) = np$

So

$E(X) = 20*0.2037 = 4.074$

The expected number of Chrome users is 4.074.

d.For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

We have that, for a binomial experiment, the variance is

$Var(X) = np(1-p)$

So

$Var(X) = 20*0.2037*(0.7963) = 3.2441$

The variance for the number of Chrome users is 3.2441.

The standard deviation is the square root of the variance. So

$\sqrt{Var(X)} = \sqrt{3.2441} = 1.8011$

The standard deviation for the number of Chrome users is 1.8011.

6 0

2 years ago