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2 years ago

What is the value of the expression below when x=5?

Mathematics

2 answers:

snow_lady [41]2 years ago

6 0

Answer:

790

Step-by-step explanation:

You put 5 instead of x and you calculat

6*5^3 +8*5= 750+40=790

Ilia_Sergeevich [38]2 years ago

5 0

The answer should be 790

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Which expression is equivalent to 12 + 24 p. A 2(6+24p) B 3(9+21p) C 4(3×6p) D 6 (2+3p)

Iteru [2.4K]

Answer:

Step-by-step explanation:

A. 2*6 = 12 but 24p*2 doesn't equal 24p

B. 3*9 doesn't equal 12, and 3*21p doesn't equal 24p

C. 4*3 equals 12 and4*6p=24p

D. 6*2=12, but 6*3p doesn't equal 24p

I hope this helps!!!

6 0

2 years ago

Read 2 more answers

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

2 years ago

Given the two vertices and the centroids of a triangle, how many possible locations are there for the third vertex?

hoa [83]

Answer:

Step-by-step explanation:

The centroid is the average of the coordinates of the three vertices. If you know two vertices (A and B) and the centroid (Q), then the third vertex (C) is ...

C = 3Q -A -B

It has only one possible location.

7 0

2 years ago

An object is dropped from a small plane. As the object falls, its distance, d, above the ground after t seconds, is given by the

DedPeter [7]

The inequality can be used to find the interval of time taken by the object to reach the height greater than 300 feet above the ground is:

$d = -16t^2 + 1000$

Solution:

The object falls, its distance, d, above the ground after t seconds, is given by the formula:

$d = -16t^2 + 1000$

To find the time interval in which the object is at a height greater than 300 ft

Frame a inequality,

$-16t^2 + 1000 > 300$

Solve the inequality

Subtract 1000 from both sides

$-16t^2 + 1000 - 1000 > 300 - 1000\\\\-16t^2 > -700$

$16t^2 < 700\\\\Divide\ both\ sides\ by\ 16\\\\t^2 < \frac{700}{16}\\\\Take\ square\ root\ on\ both\ sides\\\\t < \sqrt{\frac{700}{16}}\\\\t < \pm 6.61$

Time cannot be negative

Therefore,

t < 6.61

And the inequality used is: $-16t^2 + 1000>300$

6 0

2 years ago

Three times the difference between eighteen and thirteen

olchik [2.2K]

3*(18-13)
3* 5
15

The answer is 15

8 0

3 years ago