F and g are functions such that

Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th

Alik [6]

Answer:

$P(X$

And for this case we can use the cumulative distribution function given by:

$P(X\leq x) =\frac{x-a}{b-a}, a \leq x \leq b$

And using this formula we have this:

$P(X$

Then we can conclude that the probability that your bid will be accepted would be 0.41

Step-by-step explanation:

Let X the random variable of interest "the bid offered" and we know that the distribution for this random variable is given by:

$X \sim Unif( a= 10100, b =14700)$

If your offer is accepted is because your bid is higher than the others. And we want to find the following probability:

$P(X$

And for this case we can use the cumulative distribution function given by:

$P(X\leq x) =\frac{x-a}{b-a}, a \leq x \leq b$

And using this formula we have this:

$P(X$

Then we can conclude that the probability that your bid will be accepted would be 0.41

8 0

2 years ago

Write a rule for a transformation

lys-0071 [83]

Over the x-axis: (x, y)  (x, –y)
Over the y-axis: (x, y)  (–x, y)
Over the line y = x: (x, y)  (y, x)
Through the origin: (x, y)  (–x, –y)

6 0

2 years ago

1.)Find the height of a rectangular prism if the surface area is 148 cm2, the width is 4 cm and the length is 5 cm.

Contact [7]

1) surface of a rectangular prism=2(length x width)+2(length x height)+2(width x height)

Therefore:
148 cm²=2(5 cm x 4 cm)+2(5 cm x h)+2(4 cm x h)=
148 cm²=40 cm²+10 cm h+8 cm h
18 cm h=148 cm²-40 cm²
18 cm h=108 cm²
h=108 cm² / 18 cm=6 cm.

answer: height=6 cm

2)
Volume of a rectangular prism= length x width x height
therefore:
34 cm³=(1.7 cm)(0.5 cm) h
0.85 cm² h=34 cm³
h=34 cm³/0.85 cm²
h=40 cm.

answer: height=40 cm

3)
volume of a cylinder: πr²h
therefore.
118.79 ft³=πr²(5 ft)
r=√(118.79 ft³/5π ft)≈2.75 ft

answer: radius=2.75 ft

4)

Surface area of the pyramid with square base=4(A side)+A base
A side=(1/2)(8ft)(12 ft)=48 ft²
A base=(8 ft)(8 ft)=64 ft²

surface area=4(48 ft²)+64 ft²=256 ft²

Answer: the surface area of this pyramid would be 256 ft².

5)
surface of a cone=πrs+πr²
therefore:
radius=diameter/2=6.2 ft/2=3.1 ft
63.3 ft²=π(3.1 ft) s+π(3.1 ft)²
3.1π ft s=33.109 ft²
s=33.109 ft² /3.1π ft
s≈3.4 ft

Answer: the slant height would be 3.4 ft.

6)
volume of a square pyramid=(area of base x heigth)/3
therefore:
area of base=(6 ft)(6 ft)=36 ft²
126.97 ft³=36 ft² h /3
h=126.97 ft³/12 ft²=10.58 ft

answer: the height would be 10.58 ft.

7)
volume of a cone =(base x height)/3
base of a cone=πr²
therefore:
199.23 cm³=πr²(9 cm)/3
r=√(199.23 cm³ / 3π cm)≈4.6 cm

answer: the radius would be 4.6 cm.

5 0

2 years ago

A soccer field is a rectangle 48 meters wide and 55 meters long. The coach asks players to run from one corner to the corner dia

Sindrei [870]

Answer:

Distance = 73 m

Step-by-step explanation:

Given that,

The length of a rectangle = 48 m

The width of a rectangle = 55 m

We need to find the diagonal distance. We can use the Pythagoras theorem to find it.

$d^2=48^2+55^2\\\\d=\sqrt{5329}\\\\d=73\ m$

So, the required distance is equal to 73 m.

5 0

2 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