Which system of equations is represented by the matrix below?

The path of a breaching ocra whale can be modeled by the function k(x) = x^2 - 2x - 8, where x is the time elapsed in seconds un

____ [38]

Answer:

The time duration of the ocra whale spends under water is 4 seconds

Step-by-step explanation:

The equation that model the path of the ocra whale is presented as follows;

k(x) = x² - 2·x - 8

Where;

x = The time elapsed in seconds under water

k(x) = The height in feet below water

Therefore, the point where the ocra whale enters the water, k(x) = 0

At the point of submerging below the water surface, we have;

k(x) = 0 = x² - 2·x - 8

Factorizing the quadratic equation gives;

0 = x² - 2·x - 8 = (x - 4)·(x + 2)

Therefore, when k(x) = 0, either;

(x - 4) = 0, which gives, x = 4

(x + 2) = 0, which gives, x = -2

Given that 'x' represent the time, the time elapsed while the whale is under water, x = 4 seconds

3 0

3 years ago

A square tabletop has an area of

tiny-mole [99]

s = 3x - 15 is the required expression for perimeter of table top

Perimeter of square tabletop is 240 cm

Solution:

A square tabletop has an area given as:

$(9x^2 - 90x+225) cm^2$

The dimensions of the tabletop have the form cx - di ,where cand d are whole numbers

To find perimeter of tabletop when x = 25 centimeters

Let us first find the length of each side of square

Given area is:

$area = (9x^2 - 90x+225)$

We know that,

$area = (side)^2 = s^2$

Therefore,

$s^2 = (9x^2 - 90x+225)\\\\s^2 = (3x - 15)(3x - 15)\\\\s^2 = (3x - 15)^2$

Taking square root on both sides,

s = 3x - 15

The above expression is the required expression for perimeter of table top

To find perimeter when x = 25 centimeter

The perimeter of square is given as:

$perimeter = 4s$

perimeter = 4(3x - 15)

Substitute x = 25

perimeter = 4(3(25) - 15)

perimeter = 4(60) = 240

Therefore perimeter of square tabletop is 240 cm

4 0

3 years ago

5a - 3a + 2a simplify expression

zalisa [80]

4a is the answer I believe you’re looking for

5 0

3 years ago

Read 2 more answers

Follow the rule to find the next three numbers in pattern . describe the pattern using the terms even and odd. add 5: 1, 6, 11

vfiekz [6]

Answer:

The next 3 numbers in the pattern would be 16, 21, and 26

Step-by-step explanation:

You find this by simply adding 5 to the number before it.

4 0

3 years ago

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

B = bugs are present in a computer program.

D = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event E given that another event X has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago