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Svet_ta [14]
4 years ago
14

.

Mathematics
1 answer:
inna [77]4 years ago
7 0

Answer:

3  52/100  or  3.52

Step-by-step explanation:

three = 3

fifty two=52

hundredths=out of /100

fifty two hundredths =52/100=0.52

3.52 or 3 52/100

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A sample of bacteria was placed in a petri dish. The number of bacteria changes exponentially over time.
vredina [299]

Answer: 50

Step-by-step explanation:

50 bacteria

4 0
3 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:

<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 the value of x'.<br> (7x - 22)<br> (3x+ 2)<br> X=<br><br> Need help please and quick
mixer [17]

Answer:

X=6

Step-by-step explanation:

5 0
3 years ago
I need help can someone help me (:
vladimir2022 [97]
Hey there!

To solve the first problem, I've found it easiest to solve the equation for, say, values –2 through +2 and create a table of values for you to begin graphing this function. You may need to do more depending on the equation itself. 

Some points are: (–2, 0.75), (–1, 1.5), (0, 3), (1, 6) and (2, 12). You can check which graph matches up with these points the closest to get your answer of D. 

To solve the second problem, you'll need to use the distance equation.

x1 = –4, y1 = 3
x2 = –1, y2 = 1 
  ___________________
√ (x2–x1)^2 + (y2–y1)^2
  _________________
√ (–1–(–4)^2 + (1–3)^2
  _______________
√ (–1+4)^2 + (–2)^2
  ____________
√ (3)^2 + (–2)^2
  _____
√ 9 + 4
  ___
√ 13, making your answer D

For your third question, I always just counted the number of units the point was from the line of reflection. You'll count twice diagonally towards the line from point C for this one, staying on the "crosshairs" of the graph. All you need to do then is count two diagonal units along the same line, then you'll get your answer of (2, 6), or D. 

For your final question, A and B are immediately out, since they won't be parallel to the 4x equation. You'll need to solve both of your remaining equations for y with 2 plugged in for x; whichever one equals 7 will be your answer. In this case, it will be D. 

Hope this helped you out! :-)
3 0
3 years ago
On a coordinate plane, DEF has vertices D(1,2), E(7,2)and F(1,9). What is the distance between point E and F?
notka56 [123]

Step-by-step explanation:

Using distance formula,

Let the distance between point E and F = ef,

ef  =  \sqrt{{(7 - 1)}^{2}  +  {(2 - 9)}^{2} }  \\ ef =  \sqrt{85}  \\  = 9.219544457

8 0
3 years ago
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