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adell [148]
4 years ago
14

Before getting to school, Nancy has a few errands to run. Nancy has walk 10 blocks to the gallery, and 4 blocks to the museum, b

efore walking the final 5 blocks to arrive at school. If Nancy has already walking 11 blocks, how many blocks much she walk before arriving school
Mathematics
1 answer:
choli [55]4 years ago
6 0
If she’s walking 10 to the gallery plus 9 in total passing the museum to get to the school then that’s 19 blocks to get there. If she’s already walking 11 then she only has 8 left to go
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How to start, there is also a second part to it but I'm pretty sure once I get the right formula I will be all set
Furkat [3]
C=.35x + 9.50

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hope this helps
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3 years ago
Read 2 more answers
Brad owns a successful corporation that has a substantial earnings and profits. During the year, the following payments were mad
Bad White [126]

Answer:

That's great for Brad, wait are you talking about the Brad from clash royale? if so im impressed

Step-by-step explanation:

3 0
3 years ago
Two functions are given below: f(x) and h(x). State the axis of symmetry for each function and explain how to find it.
andrew-mc [135]

Answer:

f(x):

x =  - 4

g(x)

x = 1

Step-by-step explanation:

When a function is of the form:

f(x) = a {(x - h)}^{2}  + k

The equation of the axis of symmetry is

x = h

The given function is

f(x) = 3 {(x + 4)}^{2}  + 1

We can see that h=-4

Hence the equation of axis of symmetry is

x =  - 4

From the graph, the axis of symmetry of h(x) is the vertical line that divides the parabola into two congruent halves.

This line is

x = 1

4 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
Which of the following is the best definition of probability
SOVA2 [1]
The probability<span> of an event is the measure of the chance that the event will occur as a result of an experiment.... :)</span>
5 0
3 years ago
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