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2 years ago

How many triangles can you make with two side lengths of 6 inches that meet at a 50 degree angle.

Mathematics

2 answers:

Sonbull [250]2 years ago

5 0

<h3>Answer: Exactly One Triangle</h3>

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Explanation:

Recall that by the SAS (side angle side) congruence theorem, if we know that two pairs of sides are the same length and the angles between them are also congruent, then we can conclude the triangles are congruent. This further means that we can only construct exactly one triangle with two side lengths of 6 inches, and they form a 50 degree angle between them. Applying transformations such as rotations, translations, or reflections will not change the triangle.

Side note: This triangle is isosceles because two sides are the same length. It is not equilateral because we would need to have all three angles to be 60 degrees.

kirza4 [7]2 years ago

5 0

Answer:

One triangle only

Step-by-step explanation:

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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

Solve for m. -19m − 19 = 4m − 4m + 19

dangina [55]

-19m-19 = 4m-4m+19
-19m-19 = 0+19
-19m = 19+19
-19m = 38
m = 38÷(-19)
m = -2

7 0

3 years ago

Every evening Jenna empties her pockets and puts her change in a jar. At the end of the week she counts her money. One week she

Sergeeva-Olga [200]

A) .10 d + .25 q = 7.75
B) d + q = 40

Multiplying B) by -.10
B) -.10d -.10q = -4.0
Then adding this to A)
A) .10 d + .25 q = 7.75
.15q = 3.75

Quarters = 25
Therefore, dimes = 15
***************************************************
Double-Check
A) .10 d + .25 q = 7.75
A) .10 * 15 d + .25 * 25 = 7.75
A) 1.50 + 6.25 = 7.75

Correct!!

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3 years ago

Claire traveled 701 miles. She drove 80 miles every day. On the last day of her trip she only drove 61 miles. Write and solve an

mina [271]

Answer:

Claire traveled for 9 days.

Step-by-step explanation:

Given:

Total Distance traveled = 701 miles

Distance traveled each day = 80 miles

Distance traveled on last day = 61 miles

We need to find the number of days Claire traveled.