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3 years ago

7. In triangle ABC, side AB is 2 cm shorter than side

Mathematics

1 answer:

Mamont248 [21]3 years ago

3 0

Answer:

AC = 21, AB = 19, and BC = 22

Step-by-step explanation:

If x is the length of AC, then:

AB = x − 2

BC = x + 1

The perimeter is the sum of all three sides:

62 = x + x − 2 + x + 1

62 = 3x − 1

63 = 3x

x = 21

Therefore:

AC = 21, AB = 19, and BC = 22

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4 years ago

Suppose that p(x) is the density function for heights of American men, in inches, and suppose that p(69)=0.19. Think carefully a

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Answer:

15,2% of American men are between 68.6 and 69.4 inches tall.

Step-by-step explanation:

p(69)=0.19 tells you that 19% of all American men are 69in tall. Now, we want to know what percentage of those men are exactly between 68.6 and 69.4 inches tall. We only need to multiply the given density function by the gap between the two heights we are looking for, so:

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4 years ago

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a) ≠
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3 years ago

Write a piecewise-defined function that represents the graph?

kodGreya [7K]

Answer:

f(x) = { x, -5 ≤ x < 0 and 3x - 3, 1 ≤ x < 5

Step-by-step explanation:

The equation of one line is y = x

The equation of the other line is y = 3x - 3

Altogether,

f(x) = { x, -5 ≤ x < 0 and 3x - 3, 1 ≤ x < 5

4 0

3 years ago

Five computer program modules are ranked as M1, M2, M3, M4, and M5 according to the ascending order of effort required to debug

gulaghasi [49]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Technician selects three out of 5 systems

In C(5,3)=10ways, this can be achieved

In part a:

Space sample chooses 3 of a 5 systems

$(M_1, \ M_2,\ M_3),$ $(M_1,M_2,M_4) \ (M_1,M_2,M_5) \ (M_1,M_3,M_4) \ (M_1,M_3,M_5),$ $(M_1,M_4,M_5) \ (M_2,M_3,M_4)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5)}$

In point b:

A =MODULE WHICH INCLUDE M1 minimal amount of effort

Outcomes probable =

$(M_1,M_2,M_3),\ (M_1,M_2,M_4) \ (M_1,M_2,M_5)\ (M_1,M_3,M_4)\\\\(M_1,M_3,M_5),\ (M_1,M_4,M)5)\ =\ 6$

$\to p(A)=\frac{6}{10}\\\\$

$=0.6$

In point c:

B = highest effort that is $M_5$

Potential result=

$(M_1,M_2,M_5) \ (M_1,M_3,M_5) \ (M_2,M_3,M_5)\(M_2,M_4,M_5) \\ (M_2,M_4,M_5), \ (M_3,M_4,M_5) \ =\ 6 \\\\$

$\to B= \frac{6}{10} \\\\$

$=0.6$

$\to P(B)=10$

In point d:

$\to \ A \ intersection \ B=(M_1,M_2,M_5), \ (M_1,M_3,M_5) \ ,(M_1,M_4,M_5)$

$\to A (A \ intersection \ B) = \frac{3}{10} \\\\\ \ \ \ \ \$

$=0.3$

In point e:

$\to (A \cup B) = (M_1,M_2,M_3),\ (M_1,M_2,M_4)\ (M_1,M_2,M_5)(M_1,M_3,M_4)\ (M_1,M_3,M_5), \\ (M_1,M_4,M_5)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5) \ = \ 9$ $\to P(A \cap B)=\frac{9}{10}$

$= 0.9$

In point f:

$\to (A\cap B) = \frac{3}{10}$

$= 0.3$

In point g:

$\to (A \cup B) = \frac{7}{10}$

$=0.7$

In point h:

$\to p(A \cap B) = 0.3 \neq 0$

8 0

4 years ago