Given W1 = -2 + 3i and W2 = 4 – 4i, which complex number can be added to wi to produce wz?

The Newtown Telephone Company advertises that if you use their 10-10 service, you will always pay only 8¢ per minute, no matter

weqwewe [10]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the information above :

A) Cost equation for Newton Telephone company :

y = 8x

Where x = number of minutes used

y = total cost for x minutes

Cost equation for Tryus telephone company :

y = connection fee + (fee per minute * number of minutes)

y = 42 + (2x)

Where x = number of minutes used

y = total cost for x minutes

Amount is in cent

3 0

2 years ago

In a certain Algebra 2 class of 28 students, 11 of them play basketball and 13 of them play baseball. There are 11 students who

leonid [27]

Probability of a student playing both basketball and baseball is 7/28

Step-by-step explanation:

Step 1:

It is given the class has 28 students out of which 11 play basketball and 13 play baseball. It is also given that 11 students play neither sport.

Total number of students = 28

Students playing neither sport = 11

Students playing at least one sport = 28 - 11 = 17

Step 2:

Let N(Basketball) denote the number of students playing basketball and N(Baseball) denote the number of people playing baseball.

Then N(Basketball U Baseball) denotes the total number of students playing basketball and baseball and N(Basketball ∩ Baseball) denotes playing both basketball and baseball.

Since the number of students playing at least one sport is 17, N (Basketball U Baseball) = 17.

N (Basketball U Baseball) = N(Basketball) + N(Baseball) - N(Basketball ∩ Baseball)

N(Basketball ∩ Baseball) = N(Basketball) + N(Baseball) - N (Basketball U Baseball)

N(Basketball ∩ Baseball) = 11 + 13 - 17 = 7

Step 3:

Number of students playing both basketball and baseball = 7

Total number of students = 28

Probability of a student playing both basketball and baseball is 7/28

Step 4:

Answer:

Probability of a student playing both basketball and baseball is 7/28

3 0

2 years ago

An experiment consist of rolling two fair number cubes. The diagram shows the sample space of all equally likely outcomes. Find

Tema [17]

The answer Is D ....

5 0

3 years ago

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

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

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

So

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0

2 years ago

Write 2.3181818... as a mixed number

JulsSmile [24]

Let x= 2.3181818...

then

10x=23.181818... and

1,000x=2,318.1818...

1,000x-10x=2,318.1818...-23.181818...

990x = 2,318 - 23
990x = 2295

x=2295/990

dividing by 5 both numerator and denominator:

x= 459/198

now, x= 198 + 198+ 63

so $x= 2\frac{63}{198}$

Answer: $x= 2\frac{63}{198}$

7 0

2 years ago