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3 years ago

Read the excerpt from act 1 of The Monsters Are Due on Maple Street.

Mathematics

2 answers:

Vika [28.1K]3 years ago

5 0

Answer:

it is D. the murmurs of a group

Step-by-step explanation:

Alisiya [41]3 years ago

4 0

Answer:

Step-by-step explanation:

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Find the common ratio of the geometric sequence: 12.5,−62.5,312.5,−1562.5,…

bulgar [2K]

Answer:

-5

Step-by-step explanation:

its what you multiply by to get the next number

7 0

3 years ago

Helppp plzzzzz mathhh

spayn [35]

Answer:

A) f^-1(x)=(x-8)^3+2

Step-by-step explanation:

To find the function inverse, switch the x with y and solve for y.

$y=\sqrt[3]{x-2}+8 \\\\x=\sqrt[3]{y-2}+8\\\\(x-8)^3 = y-2\\\\ (x-8)^3 + 2 = y$

5 0

3 years ago

Determine whether line segment FG is tangent to circle E.

liubo4ka [24]

The tangent meets the radius at a right angle.

We use (the converse of) the Pythagorean Theorem to check:

$26^2 + 17^2 = 965 \ne 30^2$

Not a right triangle

Answer: FALSE

4 0

2 years ago

The stadium can seat 15,000 fans. If the stadium is 85 percent full, how many tickets were sold?

Vera_Pavlovna [14]

12,750 tickets were sold. 0.85*15,000

8 0

3 years 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

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

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

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

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

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

$P( X \le 4 ) = 0.0054$

4 0

3 years ago