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weqwewe [10]
3 years ago
10

The goat population on a farm is decreasing at a rate of 7.25% each year. If there were 75 goats originally, how many goats will

be on the farm in 4 years?
Mathematics
1 answer:
Natasha2012 [34]3 years ago
3 0

Answer:

53

Step-by-step explanation:

x/75=7.25/100

Cross Multiply

100x = 543.75

Simplify

x=5.4375

Multiply By The Number Of Years

5.4375 x 4 = 21.75

Subtract From The Original Number

75-21.75 = 53.25

<h2>Round it down (You cant have a quarter o a sheep. It would be dead)</h2><h2>53</h2>
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I Need Help!!<br><br><br> Identify each congruence transformation that maps ABC to DEF
timama [110]

Answer:

A and F

Step-by-step explanation:

Option A works as a transformating, and option a is the same as potion f

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:

<em>B</em> = bugs are present in a computer program.

<em>D</em> = 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 <em>E </em>given that another event <em>X</em> 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
Find values for the the entries in the second matrix to satisfy this equation:
Veseljchak [2.6K]

Answer:

Step-by-step explanation:

number of columns of A=number of rows of B

number of rows of A=number of rows of C

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Find the area of a triangle whose two sides are 18cm and 10cm and the perimeter is 42cm
kobusy [5.1K]

Ur answer will be - A≈69.65cm².  

<u><em>Hope this helps!!</em></u>

5 0
3 years ago
Help please asap I really need help
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