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3 years ago

HELPPPPPPPPPPPPPPPP 6TH GRADE MATH AP EX HELPPPPPPPPPPPPPPPPPPP

Mathematics

2 answers:

PIT_PIT [208]3 years ago

6 0

The answer is A, 42 cubic units

madreJ [45]3 years ago

3 0

The answer is a ; you would use length x width x height or base x height so 8 * 1 1/2 * 3 1/2 = 42

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The boys kept adding sand, and ten minutes later the volume of the sand cone had increased by 30%. What is the exact volume of t

RideAnS [48]

Answer:

a) 84π b) 109.2π

Step-by-step explanation:

The formula to find this is v=πr^2h/3

r=d/2. so the radius (r) is 6 because you have to divide the diameter (12) by 2. and the height is 7. so, 6^2=36 and 36x7=252.

252/3=84

and because the question is asking for the EXACT value, you need to just put the pi symbol next to it. Final answer for a is 84π

B) you need to find 30% of 84, which is 25.2 and add it to 84, and you will get 109.2! Final answer is 109.2π.

5 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

Simplify. (-6x^2 + 2y - 1) + (2x^2-5y + 3) A) -4x^2 + 3y -2 B) -4x^2 - 3y - 2 C) -4x^2 + 3y +2 D) -4x^2 -3y +2

Rasek [7]

Combine like terms

-6x^2+2y + -1 +2x^2 + -5y +3

( -6x^2+2x^2)+(2y-5y)+(-1+3)

= -4x^2+3y+2

Answer : C

I hope that's help !

4 0

3 years ago