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3 years ago

Find the greatest common factor of 18 and 34.

Mathematics

2 answers:

Morgarella [4.7K]3 years ago

4 0

2 is the GCF of 18 and 34

Mice21 [21]3 years ago

3 0

Answer:

Greatest common factor (GCF) of 18 and 34 is 2.

Step-by-step explanation:

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If ∆ABC = ∆EDF where the coordinates of A(0,2), B(2,4), and C(2,-1), what is the measure of DF?A-3B-3.1C-5D-5.9Please respond qu

aev [14]

The triangles ABC and EDF are congruent, meaning they have the same side lengths and angles measures.

The measure of DF, as both triangles are congruent, is equal to the measure of BC.

We can calculate the length of BC using the distance formula:

$\begin{gathered} D=\sqrt[]{(x_c-x_b)^2+(y_c-y_b_{})^2} \\ D=\sqrt[]{(2-2)^2+(-1-4)^2} \\ D=\sqrt[]{0^2+(-5)^2} \\ D=\sqrt[]{(-5)^2} \\ D=|-5| \\ D=5 \end{gathered}$

As BC is congruent with DF and BC=5, the length of DF is 5 units.

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1 year ago

Diane's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Diane

Shalnov [3]

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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:

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.

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3 years ago

A technician is launching fireworks near the event of a show. Of the remaining 11 fireworks, 4 are blue and 7 are red. If she la

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3 years ago

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Lera25 [3.4K]

It can be 2/1 I think lol

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