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2 years ago

Pls hurry and helppp NO LINKS

Mathematics

2 answers:

yuradex [85]2 years ago

8 0

Answer:

-x = 14

is not equivalent to x + 10 = 2x - 4

hope it helps

natita [175]2 years ago

8 0

THE OTHER GUY WAS CORRECT!!!but i hate links so much like ppl really spend time out of their day just to ruin ppls day that really need help on their work. so sad.

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10 POINTS FOR GRAPHING!!

forsale [732]

The answer is B. 2

The explanation is pretty simple. You look at the grids, the line and the height.

8 0

2 years ago

Simplify. 1/4y+1 1/ 2+2−1 3/4y−12

Kamila [148]

Y=-1.5

hope this helps
:)

6 0

2 years ago

Read 2 more answers

Help quick it due! Please hurry lol

snow_lady [41]

Answer:

4 km

Step-by-step explanation:

2/3(6) = 4 km

7 0

2 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

Can Some one please help me on these math questions I'll give 20 points please

MArishka [77]

Answer:

☐ -2 < 2x + 4

☐ -3x - 2 > 5

Step-by-step explanation:

-3x - 2 > 5

+ 2 + 2

___________

-3x > 7

___ ___

-3 -3

x < -2⅓ [Anytime you divide or multiply by a negative, reverse the inequality symbol.]

-2 < 2x + 4

-4 - 4

___________

-6 < 2x

__ ___

2 2

-3 < x

If you plug in -3 for x < -2⅓, you will see that it is a genuine statement because the more higher a negative gets, the lesser the integer will be, so in this case, -3 IS less than -2⅓.

I am joyous to assist you anytime.

** If it is not multi-select, then choose -2 < 2x + 4.

4 0

3 years ago