1+4=5 2+5=12 3+6=21 8+11=? puzzle answer

Answer:

The hourly rate for lectures is $7.33

Step-by-step explanation:

* Let's break down how to tackle the problem.

- For the level 3 course, examination hours are priced at double that of workshop hours.

- Workshop hours cost twice the rate of lecture hours.

- The total includes examination, workshop, and lecture hours.

- Examination lasts 3 hours, workshops 24 hours, and lectures 12 hours.

* Let’s denote the cost of lecture hours as $x per hour.

∴ The lectures cost $x per hour.

∵ Workshop charge is twice that of lectures

∴ Workshop hours cost 2(x) = 2x per hour.

∵ Examination fees are double that of workshop hours

∵ The workshop cost is 2x

∴ Examination fees are 2(2x) = 4x per hour.

- Combining costs for level 3 gives us the total of lecture, workshop, and examination hours.

∵ 12 hours for lectures

∵ 24 hours for workshops

∵ 3 hours for examinations

∵ Thus the total cost for level 3 = 12(x) + 24(2x) + 3(4x).

∴ Total cost for level 3 = 12x + 48x + 12x.

∵ Therefore, total cost = $528.

∴ 12x + 48x + 12x = 528.

∴ 72x = 528; hence we divide both sides by 72.

∴ x = 7.33.

∵ x represents the cost of lecture hours per hour.

∴ Therefore, the hourly price for lectures is $7.33.

Response:

the expected value of this raffle if you purchase 1 ticket = -0.65

Breakdown of the calculation:

Details:

5,000 tickets are sold at​ $1 each for a charitable raffle

Winners will be chosen at random with cash prizes as follows: 1 prize of ​$500​, 3 prizes of ​$300​, 5 prizes of ​$50​, and 20 prizes of​ $5.

Therefore, the value and its respective probability can be calculated as follows:

Value                              Probability

$500 - $1 = $499              1/5000

$300 - $1 = $299              3/5000

$50 - $1 = $49                    5/5000

$5 - $1 = $4                     20/5000

-$1                           1 - 29/5000 = 4971/5000

The expected value of the raffle when buying 1 ticket is computed as follows:

So, the expected value of this raffle when one ticket is purchased = -0.65

Question:

Which statement is accurate concerning the function’s discontinuities

A) There are gaps at x = 7 and.

B) Asymptotes exist at x = 7 and.

C) Asymptotes exist at x = –7 and.

D) Gaps are present at (–7, 0) and.

Answer:

B) Asymptotes exist at x = 7 and

Step-by-step explanation:

Given:

Goal:

Identify the correct statement

We need to factor the denominator first.

To express x in terms of (3x+4) and (x-7):

3x + 4 =

3x = -4

Divide both sides by 3:

x - 7

x = 7

Next, evaluate the limit when and at (x = 7)

lim f(x) as = ±∞

lim f(x) as (x=7) = ±∞

Since both scenarios result in the denominator approaching zero, they represent asymptotes.

Thus, asymptotes are found at and x=7

Option B is determined to be correct

Part A:

The probability that all three strangers have their birthdays on a Wednesday is calculated as





Part B:

The probability that the birthdays of the three individuals fall on distinct days throughout the week is calculated as





Part C:

The probability that none of the three have their birthdays on a Saturday is determined by