Which expression(s) have a greatest common factor (GCF) of 3xy2 with 42xy4

A school organised a concert for charity the graph and table showing information about 100 tickets sold

Inessa [12570]

Answer:

The cumulative frequency table displays the scores that some students achieved in a test. (a) Create a cumulative frequency graph based on this data on the grid provided. [2]...

Step-by-step explanation:

8 0

3 months ago

Read 2 more answers

Jane is training for a triathlon. After swimming a few laps, she leaves the health club and bikes 16 miles south. She then runs

PIT_PIT [12445]

Answer:

The formula to determine the distance Jane's trainer bikes is $x=\sqrt{16^2+12^2}$ .

Detailed explanation:

A diagram has been included for clarity.

Known values:

Distance biked to the south = 16 miles

Distance ran west = 12 miles

We need to determine the distance covered by Jane's trainer.

Solution:

Let the biking distance be represented by 'x'.

We will assume it forms a right-angled triangle.

According to the Pythagorean theorem, it states that;

"The square of the hypotenuse is equal to the sum of the squares of the other two sides."

When set in equation form, this gives us;

$x^2=16^2+12^2\\\\x=\sqrt{16^2+12^2}$

Thus, the equation representing the distance biked by Jane's trainer is $x=\sqrt{16^2+12^2}$ .

Upon solving, we find;

$x=\sqrt{16^2+12^2}\\\\x= \sqrt{256+144}\\ \\x=\sqrt{400} \\\\x=20\ miles$

Consequently, Jane's trainer covers a distance of 20 miles.

7 0

1 month ago

According to a survey by Bankrate, of adults in the United States save nothing for retirement (CNBC website). Suppose that adult

tester [12383]

Complete Question

The complete question appears in the first uploaded image

Answer:

a) Yes, selecting 15 corresponds to a binomial experiment

b)

c)

d) $P(r = 15) = 3.2768 *10^{-11}$

Step-by-step explanation:

Regarding question a:

For an experiment to qualify as binomial

the trials have to be independent

each trial must yield one of two possible outcomes

Given that the selection of 15 individuals is random, we ascertain that the trials are independent and the outcomes are “either the individual saves for retirement or does not save for retirement.”

Therefore, we conclude that the selection of 15 people at random is indeed a binomial experiment.

In question b:

The probability that all selected adults do not save for retirement is mathematically modeled as

$P(r = n) = ^nC_r * p^r * q^{n-r}$

Here C signifies combination

r = 15 implies all selected adults

n refers to the population size equating to 15

From the problem, p = 0.20

and q can be calculated as

=>

=> $q = 1 - p$

Thus

$P(r = 15) = ^{15}C_{15} * p^{15} * q^{15-15}$

$P(r = 15) = 3.2768 *10^{-11}$ Regarding question c:

The probability that exactly five of the selected adults do not save for retirement is mathematically modeled as

$P(r = 5) = ^{15} C_5 * (0.20)^5 * (0.80)^{15}$

$P(r = 5) = 0.1032$

In relation to question d:

The probability that at least one of the selected adults opts not to save for retirement can be mathematically expressed as

$P(r \ge 1 ) = 1 - P (r = 0 )$

$P(r \ge 1 ) = 1 - [ ^{15} C _ 0 * (0.20)^{0} * (0.80 )^{15}]$

$P(r \ge 1 ) = 1 - 0.0352$

$P(r \ge 1 ) = 0.9648$

4 0

2 months ago

Using the extended Euclidean algorithm, find the multiplicative inverse of a. 1234 mod 4321 b. 24140 mod 40902

AnnZ [12381]

(a) The multiplicative inverse of 1234 (mod 4321) is x so that 1234*x ≡ 1 (mod 4321). We can apply Euclid's algorithm:

4321 = 1234 * 3 + 619

1234 = 619 * 1 + 615

619 = 615 * 1 + 4

615 = 4 * 153 + 3

4 = 3 * 1 + 1

Now we will express 1 as a linear combination of 4321 and 1234:

1 = 4 - 3

1 = 4 - (615 - 4 * 153) = 4 * 154 - 615

1 = 619 * 154 - 155 * (1234 - 619) = 619 * 309 - 155 * 1234

1 = (4321 - 1234 * 3) * 309 - 155 * 1234 = 4321 * 309 - 1082 * 1234

This reduces to

1 ≡ -1082 * 1234 (mod 4321)

Thus, the inverse is

-1082 ≡ 3239 (mod 4321)

(b) Since both 24140 and 40902 are even, their GCD cannot equal 1, indicating no inverse exists.

8 0

3 months ago

for the level 3 course, examination hours cost twice as much as workshop hours and workshop hours cost twice as much as lecture

Leona [12618]

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.

6 0

3 months ago