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3 years ago

Help plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

Mathematics

1 answer:

svetlana [45]3 years ago

8 0

Answer:

2,250.

Step-by-step explanation:

Add all of the bills together. It says the minimum, so in order to pay all of the bills, you'll need 2,250.

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Given the triangle that has a base of five less than triple the height. The height is 2x+6.

Deffense [45]

Answer:

Step-by-step explanation:

H = 2x+6 - the hight

3H = 3(2x+6) - triple the hight

3H-5 = 3(2x+6) - 5 - five less than triple the height

Area of triangle: A = 0.5BH

B = 3(2x+6)-5 = 6x + 18 - 5 = 6x + 13

H = 2x+6

So:

A = 0.5(2x+6)(6x+13) = (x+6)(6x+13) = 6x² + 13x + 36x + 78

A = 6x² + 49x + 78

5 0

3 years ago

(pls ignore this. XXXXX)

Gelneren [198K]

Answer:

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

6 0

2 years ago

Read 2 more answers

A medical researcher needs 37 people to test the effectiveness of an experimental drug. If 101 people have volunteered for the t

Rashid [163]

Answer:

The number of ways to select 37 people from 101 is, 5,397,234,129,638,871,133,346,507,775.

Step-by-step explanation:

In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.

The formula to compute the combinations of k items from n is given by the formula:

${n\choose k}=\frac{n!}{k!\cdot (n-k)!}$

Compute the number of ways to select 37 people from 101 as follows:

${101\choose 37}=\frac{101!}{37!\cdot (101-37)!}$

$=\frac{101!}{37!\times 64!}\\\\=\frac{101\times 100\times 99\times....64!}{37!\times 64!}\\\\=5397234129638871133346507775$

Thus, the number of ways to select 37 people from 101 is, 5,397,234,129,638,871,133,346,507,775.

5 0

2 years ago

The problem of matching aircraft to passenger demand on each flight leg is called the flight assignment problem in the airline i

Darina [25.2K]

Answer:

Step-by-step explanation:

Given that the demand for the 6 p.m. flight from Toledo Express Airport to Chicago's O'Hare Airport on Cheapfare Airlines is normally distributed with a mean of 132 passengers and a standard deviation of 42

Let X be the no of passengers who report

X is N(132, 42)

Or Z is $\frac{x-132}{42}$

a) Suppose a Boeing 757 with a capacity of 183 passengers is assigned to this flight.

the probability that the demand will exceed the capacity of this airplane

= $P(X>183) = P(Z>1.21) =$

$=0.5-0.3869\\=0.1131$

b) the probability that the demand for this flight will be at least 80 passengers but no more than 200 passengers

= $P(80\leq x\leq 200)\\= P(-1.23\leq z\leq 1.62)\\$

=0.4474+0.3907

=0.8381

c) the probability that the demand for this flight will be less than 100 passengers

$=P(x$

d) If Cheapfare Airlines wants to limit the probability that this flight is overbooked to 3%, how much capacity should the airplane that is used for this flight have? passengers

= $P(Z>c)=0.03\\c=1.88\\X=132+1.88(42)\\=210.96$