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3 years ago

WILL MARK BRAINLIEST PLEASE HELP!

Mathematics

1 answer:

Allushta [10]3 years ago

7 0

The Answer is x= -6 and x= -6

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BabaBlast [244]

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3 years ago

Check-Ins 2 NC Math 1

fomenos

The maximum amount of profit the carnival makes based on the number of tickets sold is 6.5 thousand of dollars,

<h3>How to determine the difference?</h3>

The function is given as:

f(x) = -0.5x^2 + 5x - 6

Differentiate the function

f'(x) = -x + 5

Set to 0

-x + 5 = 0

Make x the subject

x = 5

Substitute x = 5 in f(x)

f(5) = -0.5 * 5^2 + 5 * 5 - 6

f(5) = 6.5

The table is not given.

Hence, the maximum amount of profit the carnival makes based on the number of tickets sold is 6.5 thousand of dollars,

Read more about quadratic functions at:

brainly.com/question/18797214

5 0

1 year ago

At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more

I am Lyosha [343]

Answer:

The probability is $P( X \le 4 ) = 0.0054$

Step-by-step explanation:

From the question we are told that

The percentage that are on time is p = 0.76

The sample size is n = 11

Generally the percentage that are not on time is

$q = 1- p$

$q = 1- 0.76$

$q = 0.24$

The probability that no more than 4 of them were on time is mathematically represented as

$P( X \le 4 ) = P(1 ) + P(2) + P(3) + P(4)$

=> $P( X \le 4 ) = \left n } \atop {}} \right.C_1 p^{1} q^{n- 1} + \left n } \atop {}} \right.C_2p^{2} q^{n- 2} + \left n } \atop {}} \right.C_3 p^{3} q^{n- 3} + \left n } \atop {}} \right.C_4 p^{4} q^{n- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{11- 1} + \left 11 } \atop {}} \right.C_2p^{2} q^{11- 2} + \left 11 } \atop {}} \right.C_3 p^{3} q^{11- 3} + \left 11 } \atop {}} \right.C_4 p^{4} q^{11- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{10} + \left 11 } \atop {}} \right.C_2p^{2} q^{9} + \left 11 } \atop {}} \right.C_3 p^{3} q^{8} + \left 11 } \atop {}} \right.C_4 p^{4} q^{7}$

$= \frac{11! }{ 10! 1!} (0.76)^{1} (0.24)^{10} + \frac{11!}{9! 2!} (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!} (0.76)^{3} (0.24)^{8} + \frac{11!}{7!4!} (0.76)^{4} (0.24)^{7}$