The probability any one system works is 0.99
So the probability of any one system failing is 1-0.99 = 0.01, so basically a 1% chance of failure for any one system

Multiply out the value 0.01 with itself four times
0.01*0.01*0.01*0.01 = 0.000 000 01
I'm using spaces to make the number more readable
So the probability of all four systems failing is 0.00000001
Subtract this value from 1 to get
1 - 0.00000001 = 0.99999999

The answer is 0.99999999 which is what we'd expect. The probability of at least one of the systems working is very very close to 1 (aka 100%)

7 0

3 years ago

Which expression could be used to solve the problem?

Paladinen [302]

THE ANSWER IS D whoops lol sorry i was in an argument lol

4 0

3 years ago

The larger leg of a right triangle is 2 feet longer than its smaller leg. The hypotenuse is 4 feet longer than the smaller leg.

maks197457 [2]

Hope it helps you:))

7 0

3 years ago

The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the

Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as $e=\frac{c}{a}$

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

$\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)$

b) Eccentricity =5

$5=\frac{c}{a} \:c=5a$

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

$If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0$

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

$a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)$

$Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1$

7 0

2 years ago