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Oksanka [162]
1 year ago
5

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Mathematics
2 answers:
Tema [17]1 year ago
7 0

Answer: 88

Step-by-step explanation:

88

Oduvanchick [21]1 year ago
6 0

Answer:

  see attached

Step-by-step explanation:

The Pythagorean theorem can be used to find the hypotenuse associated with each pair of legs. That tells you ...

  c² = a² +b² . . . . . legs a, b; hypotenuse c

__

<h3>alternate form of Pythagorean theorem</h3>

For the purpose of this problem, it is convenient to consider a slightly different form of the equation.

For legs √a and √b, the hypotenuse √c is given by ...

  (√c)² = (√a)² +(√b)²

  c = a +b

That is ...

  legs √a, √b ⇒ hypotenuse √(a+b)

__

<h3>application to this problem</h3>

Since the legs are (mostly) given in terms of square roots, the value under the radical for the hypotenuse is simply the sum of those:

legs: √1, √2 ⇒ hypotenuse √(1+2) = √3

legs: √2, √3 ⇒ hypotenuse √(2+3) = √5

legs: √5, √3 ⇒ hypotenuse √(5+3) = √8

legs: √5, √1 ⇒ hypotenuse √(5+1) = √6

_____

<em>Additional comment</em>

You may not see the leg lengths given as square roots very often. This is a rather unusual set of problems for hypotenuse length.

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A certain company sends 40% of its overnight mail parcels by means of express mail service A1. Of these parcels, 4% arrive after
Harrizon [31]

Answer:

(a) The probability that a randomly selected parcel arrived late is 0.026.

(b) The probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c) The probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d) The probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

Step-by-step explanation:

Consider the tree diagram below.

(a)

The law of total probability sates that: P(A)=P(A|B)P(B)+P(A|B')P(B')

Use the law of total probability to determine the probability of a parcel being late.

P(L)=P(L|A_{1})P(A_{1})+P(L|A_{2})P(A_{2})+P(L|A_{3})P(A_{3})\\=(0.04\times0.40)+(0.01\times0.50)+(0.05\times0.10)\\=0.026

Thus, the probability that a randomly selected parcel arrived late is 0.026.

(b)

The conditional probability of an event A provided that another event B has already occurred is:

P(A|B)=\frac{P(B|A)P(A)}{P(B)}

Compute the probability that a parcel was late was being shipped through the overnight mail service A₁ as follows:

P(A_{1}|L)=\frac{P(L|A_{1})P(A_{1})}{P(L)} \\=\frac{0.04\times 0.40}{0.026} \\=0.615

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

P(A_{2}|L)=\frac{P(L|A_{2})P(A_{2})}{P(L)} \\=\frac{0.01\times 0.50}{0.026} \\=0.192

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

P(A_{3}|L)=\frac{P(L|A_{3})P(A_{3})}{P(L)} \\=\frac{0.05\times 0.10}{0.026} \\=0.192

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

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3 years ago
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