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Monica [59]
3 years ago
11

Brandon is an amateur marksman. When he takes aim at a particular target on the shooting range, there is a 0.1, point, probabili

ty that he will hit it. One day, Brandon decides to attempt to hit 10 such targets in a row.
Assuming that Brandon is equally likely to hit each of the 10 targets, what is the probability that he will hit at least one of them?
Mathematics
1 answer:
kicyunya [14]3 years ago
7 0

Answer:

.65

Step-by-step explanation:

Strategy:

In this situation it is much easier to calculate the probability of the event we are looking for (he hits at least one target) by calculating the probability of its complement (he misses every target), and subtracting from 1.

In other words, we can use this strategy:

P(at least one hit)=1-P(miss all 10)

Calculations:

P(at least one hit)

=1-P(miss all 10)

=1-(0.9)^10

≈1-0.349

≈0.65

Answer:

P(at least one hit)≈0.65

I hope this helps!!!

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Find the zeros of the following polynomial functions, with their multiplicities. (a) f(x)= (x +1)(x − 1)(x² +1) (b) g(x) = (x −
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Answer:

a) zeros of the function are x = 1 and, x = -1

b) zeros of the function are x = 2 and, x = 4

c) zeros of the function are x = \frac{3}{2}

d) zeros of the function are x = \frac{-4}{3}  and, x = 17

Step-by-step explanation:

Zeros of the function are the values of the variable that will lead to the result of the equation being zero.

Thus,

a) f(x)= (x +1)(x − 1)(x² +1)

now,

for the (x +1)(x − 1)(x² +1) = 0

the condition that must be followed is

(x +1) = 0 ..........(1)

or

(x − 1) = 0 ..........(2)

or

(x² +1) = 0 ...........(3)

considering the equation 1, we have

(x +1) = 0

or

x = -1

for

(x − 1) = 0

x = 1

and,

for (x² +1) = 0

or

x² = -1

or

x = √(-1)         (neglected as it is a imaginary root)

Thus,

zeros of the function are x = 1 and, x = -1

b) g(x) = (x − 4)³(x − 2)⁸

now,

for the (x − 4)³(x − 2)⁸ = 0

the condition that must be followed is

(x − 4)³ = 0 ..........(1)

or

(x − 2)⁸ = 0 ..........(2)

considering the equation 1, we have

(x − 4)³ = 0

or

x -4 = 0

or

x = 4

and,

for (x − 2)⁸ = 0

or

x - 2 = 0

or

x = 2        

Thus,

zeros of the function are x = 2 and, x = 4

c) h(x) = (2x − 3)⁵

now,

for the (2x − 3)⁵ = 0

the condition that must be followed is

(2x − 3)⁵ = 0

or

2x - 3 = 0

or

2x = 3

or

x = \frac{3}{2}

Thus,

zeros of the function are x = \frac{3}{2}

d)   k(x) =(3x +4)¹⁰⁰(x − 17)⁴

now,

for the (3x +4)¹⁰⁰(x − 17)⁴ = 0

the condition that must be followed is

(3x +4)¹⁰⁰ = 0 ..........(1)

or

(x − 17)⁴ = 0 ..........(2)

considering the equation 1, we have

(3x +4)¹⁰⁰ = 0

or

(3x +4) = 0

or

3x = -4

or

x = \frac{-4}{3}

and,

for (x − 17)⁴ = 0

or

x - 17 = 0

or

x = 17        

Thus,

zeros of the function are x = \frac{-4}{3}  and, x = 17

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3 years ago
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