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4 years ago

What is i + 6i? You have to simplify

Mathematics

1 answer:

marissa [1.9K]4 years ago

3 0

$i=1i\\\\i+6i=1i+6i=(1+6)i=\boxed{7i}$

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a surveyor measures the distance to the two landmarks as shown in the diagram what is the distance d. between the landmarks

Inessa [10]

<h3>The distance between two landmarks is 123 meters</h3>

<em><u>Solution:</u></em>

We have to find the distance between two landmarks

<em><u>Use the law of cosines</u></em>

The third side of a triangle can be found when we know two sides and the angle between them

$c^2 = a^2 + b^2 - 2ab\ cos c$

Here, angle between 90 meters and 130 meters is 65 degrees

From figure,

a = 90

b = 130

c = d

Therefore,

$d^2 = 90^2 + 130^2 - 2(90)(130)\ cos 65\\\\d^2 = 8100 + 16900 - 23400\ cos\ 65\\\\d^2 = 25000 - 23400 \times 0.4226\\\\d^2 = 25000 - 9889.267\\\\d^2 = 15110.73\\\\d = 122.92\\\\d \approx 123$

Thus, the distance between two landmarks is 123 meters

8 0

4 years ago

USPshnik [31]

Answer:

$17,500 at 8%
$17,500 at 14%

Step-by-step explanation:

The fraction that needs to be invested at the higher rate is ...

(11% - 8%)/(14% -8%) = 3%/6% = 1/2

Half the money should be invested at each of the rates:

$17,500 for 8% return

$17,500 for 14% return.

_____

If you let x represent the amount at the higher rate, then the total return is ...

14%x + 8%(35000 -x) = 11%(35000)

(14% -8%)x = (11% -8%)(35000) . . . . subtract 8%·35000

x = (11% -8%)/(14% -8%)×35000 . . . . divide by the coefficient of x

Note that this formula is exactly the one we started with, above.

4 0

3 years ago

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

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:

$n = 12, p = 0.5$

(a) 2 or 5 children

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

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

3 years ago

Determine two numbers that have a sum of 12 but a difference of 12

sveta [45]

Answer:

0 And 12

Step-by-step explanation:

0 + 12 = 12

But 0 is 12 away from 12. hope i helped :)

6 0

3 years ago

Read 2 more answers

There are 31 dogwood trees and 22 maple trees currently in the park. the park workers will plant more dogwood trees today.When t

nalin [4]

If there was 31 dogwood trees already there and after there was a total of 96 dogwood trees therefore all you would have to do is subtract (96-31) you would get 65 so the answer is 65

8 0

4 years ago