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morpeh [17]
3 years ago
5

If f(x)=5x+a and f-1(10)=-1, find a

Mathematics
1 answer:
zhenek [66]3 years ago
4 0

Answer:

15

Step-by-step explanation:

Given the function

f(x)=5x+a

and f^{-1} (10)=-1(f^{-1} \text{is the inverse function of f})

let f(x)=y

Substitute in f(x)

y=5x+a\\y-a=5x\\x=\frac{y-a}{5}

From the definition inverse function

if f(x)=y    ⇒   f^{-1} (y)=x

Substitute this in the above expression

f^{-1} (y)=\frac{y-a}{5}

Substitute y=10

⇒

f^{-1} (10)=\frac{10-a}{5}\\-1=\frac{10-a}{5}\\-5=10-a\\a=10+5\\a=15

Therefore a=15

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What is 0.04% of 30? Round to the nearest hundredth
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Step-by-step explanation:

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Answer:

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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

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3 years ago
............................
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Answer:

............................

Step-by-step explanation:

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