Is 4 quartes greater than 10 pints

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KIM [24]

Answer:

Step-by-step explanation:

:)

4 0

2 years ago

Read 2 more answers

Find MP. MN=17 NP=3y MP=5y+9

8_murik_8 [283]

Answer:

$MP = 29$

Step-by-step explanation:

Given:

$MN = 17$

$NP = 3y$

$MP = 5y + 9$

$MN + NP = MP$ (segment addition postulate)

$17 + 3y = 5y + 9$ (substitution)

$3y = 5y + 9 - 17$ (subtracting 17 from each side)

$3y = 5y + 8$

$3y - 5y = 5y - 8 - 5y$ (Subtracting 5y from both sides)

$-2y = 5y - 8 - 5y$

$-2y = - 8$

$y = 4$ (dividing both sides by -2)

$MP = 5y + 9$

Plug in the value of y

$MP = 5(4) + 9 = 20 + 9$

$MP = 29$

4 0

3 years ago

If 3t-7=5t, then 6t =

Natalija [7]

The answer is: " -21 " .
__________________________________________
We are asked to solve for: "6t" ; which is: "6 * t" .
__________________________________________
Given: 3t − 7 = 5t ; Solve for "t" ; then solve for "6t" ;

     3t − 7 = 5t ;   Subtract "3t" from each side of the equation;
____________________________________________
    3t − 7 − 3t = 5t − 3t ;

to get: -7 = 2t ;

Divide EACH side of the equation by "2" ; to isolate "t" on one side of the equation; and to solve for "t" ;

-7/2 = 2t / 2 ;

-3.5 = t ; ↔ t = - 3.5 ;
____________________________________________
Now, solve for "6t" ;

6 t = 6*(-3.5) = - 21 .   The answer is: " -21 " .
____________________________________________

6 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

2 years ago

Help!!!!!!!!!!!!!!!!!!!

Vika [28.1K]

Answer:

base

median

hypotenuse answer of no. a b and c

6 0

2 years ago