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Bess [88]
3 years ago
11

Given: a = 9, c = 5, B = 120 b=

Mathematics
1 answer:
Zinaida [17]3 years ago
6 0
B = 120 because it is in your sum
You might be interested in
4m – 3(m + 1) – 4 < 0
m_a_m_a [10]

Answer:1m-7<0

Step-by-step explanation:

4m-3m-3-4

1m-7<0

8 0
3 years ago
The equation f = v + at represents the final velocity of an object, f, with an initial velocity, v, and an acceleration rate, a,
Elenna [48]

Answer:

(f-v)/a = t

Step-by-step explanation:

f = v + at

Subtract v from each side

f-v = v-v + at

f-v = at

Divide each side by a

(f-v)/a = at/a

(f-v)/a = t

4 0
3 years ago
Suppose that in one metropolitan area, 25% of all home- owners are insured against earthquake damage. Four home-owners are to be
emmasim [6.3K]

Answer:

Step-by-step explanation:

Given that,

25% home owners are insecure of earthquake problem

If we select 4 home owners at random

let X denote the number among the four who have earthquake insurance

Let find the probability distribution of X

Let S- denotes a home owner who has insurance

Let F denotes a home owner who does not have insurance.

Then, P(S) =25% = 0.25

P(F) = 1 — P(S) = 1 —0.25

P(F) = 0.75

The possible outcomes of X is

X(0) = If no person has insurance

X(1) = If only 1 person has insurance

X(2) = If only 2 persons has insurance

X(3) = If only 3 persons has insurance

X(4) = If only 4 persons has insurance.

The cardinality of the sample space is n(C) = 2ⁿ = 2⁴ => 16

So, the sample space is given as

{FFFF, FFFS, FFSF, FFSS, FSFF, FSFS, FSSF, FSSS, SFFF, SFFS, SFSF, SFSS, SSFF, SSFS, SSSF, SSSS}

For X=0, the possible is {FFFF} i.e. no insurance, the one without insurance.

P(X) = 0.25×0.25×0.25×0.25

P(X) = 0.25⁴

P(X) = 0.00390625

For X=1, the possible outcome are

FFFS, FFSF, FSFF, SFFF

P(X=1) = 0.25³•0.75 + 0.25³•0.75 + 0.25³•0.75 + 0.25³•0.75

P(X=1) = 0.046875

For X=2 the possible outcomes are

FFSS, FSFS, FSSF, SFFS, SFSF, SSFF,

P(X=2)=0.25²•0.75²+0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²

P(X=2) = 0.2109375

For X=3 the possible outcomes are

FSSS, SFSS, SSFS, SSSF

P(X=3) = 0.25•0.75³+0.25•0.75³+ 0.25•0.75³+0.25•0.75³

P(X=3) = 0.421875

X=4 the possible outcomes are

SSSS

P(X=4) = 0.75×0.75×0.75×0.75

P(X=4) = 0.31640625.

Or

Using normal distribution

P(X=k) = ⁿCk • 0.25^k • 0.75^(4-k)

So,

P(X=0) = 4C0 • 0.25^0 • 0.75^4

P(X=0) = 0.31640625

P(X=1) = 4C1 • 0.25^1 • 0.75^3

P(X=1) = 0.421875

P(X=2) = 4C2 • 0.25^2 • 0.75^2

P(X=2) = 0.2109375

P(X=3) = 4C3 • 0.25^3 • 0.75^1

P(X=3) = 0.046875

P(X=4) = 4C4 • 0.25^4 • 0.75^0

P(X=4) = 0.00390625.

6 0
3 years ago
A 1414​-ft ladder leans against a wall at a point 55 feet above the ground. how far is the bottom of the ladder from the​ wall?
viva [34]
Assume ladder length is 14 ft and that the top end of the ladder is 5 feet above the ground.

Find the distance the bottom of the ladder is from the base of the wall.

Picture a right triangle with hypotenuse 14 feet and that the side opposite the angle is h.  Then sin theta = h / 14, or theta = arcsin 5/14.  theta is

0.365 radian.  Then the dist. of the bot. of the lad. from the base of the wall is 

14cos theta = 14cos 0.365 rad = 13.08 feet.  This does not seem reasonable; the ladder would fall if it were already that close to the ground.

Ensure that y ou have copied this problem accurately from the original.

7 0
3 years ago
Find y, at and tu I need help please help
lesantik [10]

Answer:

there is no picture... but i will try my best to help you if you post picture in another question

8 0
2 years ago
Read 2 more answers
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