N2,3,5,7,10,21В наборе будет внутренний набор

4. Paula drew a map of her neighborhood on a coordinate grid. A grocery store is located at point (5,-4). A park is located at p

Paha777 [63]

Answer:

Midpoint Formula= (x1+x2)/2, (y1+y2)/2

Plug in the numbers

Answer= (0, 0)

I hope this helps you!!

6 0

3 years ago

Many years ago, the towns of Franklin and Chester agreed to begin posting their populations on signs just outside of their towns

iVinArrow [24]

Answer:

The answer is 9

Step-by-step explanation:

4 0

4 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

3 years ago

I need help finding if it’s either SOH, CAH or TOA and the indicated side of the triangle, thank you!

Brilliant_brown [7]

Answer:

$\huge\boxed{IJ \approx 2.01}$

Step-by-step explanation:

In order to find the side mentioned (IJ), we need to use SOH CAH TOA.

SOH CAH TOA is an acronym to help us remember what sin, cos, and tan mean. It stands for:

Sin = Opposite / Hypotenuse

Cosine = Adjacent / Hypotenuse

Tan = Opposite / Adjacent

Since we know the measure of angle K (42) and we know one of the sides, we can use this to find the missing length.

Since the side given to us is the hypotenuse, and we're looking for the side opposite of the angle (IJ), the only possible one to use would be SIN as it includes Opposite and Hypotenuse.

Our equation is now this: $\text{sin(42)} = \frac{x}{3}$