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3 years ago

56-99+567(686)/8x-56x+10

Mathematics

1 answer:

Alex17521 [72]3 years ago

6 0

Answer:

1200(-48+10)

Step-by-step explanation:

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1. f(x) = x(x-6)(x+3)

maks197457 [2]

Answer:

x3squared-3x2squared-18x

Step-by-step explanation:

1 Expand by distributing terms.

({x}^{2}-6x)(x+3)(x

−6x)(x+3)

2 Use the FOIL method: (a+b)(c+d)=ac+ad+bc+bd(a+b)(c+d)=ac+ad+bc+bd.

{x}^{3}+3{x}^{2}-6{x}^{2}-18xx

+3x

−6x

−18x

3 Collect like terms.

{x}^{3}+(3{x}^{2}-6{x}^{2})-18xx

+(3x

−6x

)−18x

4 Simplify.

{x}^{3}-3{x}^{2}-18xx

−3x

−18x

5 0

3 years ago

PLEASE HELP ME IM BEGGING

anygoal [31]

Stuck on the same thing, tried reviewing and reviewing but now I’ll just try finding the answer key

8 0

3 years ago

Plz help on #3 ASAP!:)

masya89 [10]

Answer:

Step-by-step explanation:

4 0

3 years ago

which equation represents this sentence? five more than three times the number is one-third more than the sum of the number and

lidiya [134]

3n + 5 = (n + n) + 1/3

4 0

3 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