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4 years ago

!!!! will give brainliest:

Mathematics

1 answer:

Nataly [62]4 years ago

4 0

Answer:

the domain will be all real #'s bc the arrows are continuous/never ending. The range would be y<u><</u>4 because the line starts at 4 and goes down.

Hope this helped you :)

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Maslowich

You can repot 5 plants in 30 minutes.

5•4=20
30•4=120

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4 years ago

after graduating from college Carlos received two different job offers both pay starting salary of 62,000 per year but one job p

Maru [420]

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3 years ago

Find all the zeroes of the polynomial function f ( x ) = x 3 − 5 x 2 + 6 x − 30

vitfil [10]

Answer:

5, ±√6i

Step-by-step explanation:

f(x)=x^3-5x^2+6x-30=x^2(x-5)+6(x-5)=(x^2+6)(x-5)

to make f(x)=0, x^2+6=0 or x-5=0

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3 0

3 years ago

Find all sets of four consecutive integers whose sum is between 95 and 105

Nat2105 [25]

Hello,

Let n-2,n-1,n,n+1 the foru numbers
s=n-2+n-1+n+n+1=4n-2

95<=4n-2<=105
==>97<=7n<=107
==>24.25<=n<=26.75
==>25<=n<=26
if n=25 s=98
if n=26 s=102

7 0

4 years ago

Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving

lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0

3 years ago