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Anna [14]
3 years ago
9

Twenty-two percent of the large pieces of mail that Rachel received this week were magazines and the rest were catalogs. Of the

catalogs, 36 percent were for clothing. To the nearest whole percent, what is the probability that a randomly chosen large piece of Rachel’s mail was a clothing catalog?
Mathematics
2 answers:
Marizza181 [45]3 years ago
6 0
Since 22% were not catalogs, 78% were catalogs. 36% of those were clothing:
.36 * .78 = .28 or 28%
Alinara [238K]3 years ago
3 0

Answer:

28%

Step-by-step explanation:

Given :

Twenty-two percent of the large pieces of mail that Rachel received this week were magazines and the rest were catalogs.

Of the catalogs, 36 percent were for clothing.

To Find: To the nearest whole percent, what is the probability that a randomly chosen large piece of Rachel’s mail was a clothing catalog

Solution:

Magazines = 22%=0.22

we are given that Rest are catalogs

So, catalogs = 100%-22% = 78% = 0.78

Of the catalogs, 36 percent were for clothing.

Out of catalogs Clothing = 0.36

So, probability that a randomly chosen large piece of Rachel’s mail was a clothing catalog = 0.36 \times 0.78=0.2808

To the nearest whole percent, the probability that a randomly chosen large piece of Rachel’s mail was a clothing catalog = 0.2808 \times 100=28.08\%

Hence  the probability that a randomly chosen large piece of Rachel’s mail was a clothing catalog to the nearest whole percent is 28%

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Answers:

  • Satellite is approximately <u>2446.43 km</u> from station A.
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=========================================================

Explanation:

I'm assuming tracking stations A and B are at the same elevation and are on flat ground. In reality, this is likely not the case; however, for the sake of simplicity, we'll assume this is the case.

The diagram is shown below. Points A and B describe the two stations, while point C is the satellite's location. Point D is on the ground directly below the satellite. We have these lengths

  • AB = 60 km
  • AD = x
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Focusing on triangle ACD, we can apply the tangent rule to isolate h.

tan(angle) = opposite/adjacent

tan(A) = CD/AD

tan(86.4) = h/x

x*tan(86.4) = h

h = x*tan(86.4)

We'll use this later in the substitution below.

--------------------

Now move onto triangle BCD. For the reference angle B = 85, we can use the tangent rule to say

tan(angle) = opposite/adjacent

tan(B) = CD/DB

tan(B) = CD/(DA+AB)

tan(85) = h/(x+60)

tan(85)*(x+60) = h

tan(85)*(x+60) = x*tan(86.4) .............  apply substitution; isolate x

x*tan(85)+60*tan(85) = x*tan(86.4)

60*tan(85) = x*tan(86.4)-x*tan(85)

60*tan(85) = x*(tan(86.4)-tan(85))

x*(tan(86.4)-tan(85)) = 60*tan(85)

x = 60*tan(85)/(tan(86.4)-tan(85))

x = 153.612786190499

--------------------

We'll use this approximate x value to find h

h = x*tan(86.4)

h = 153.612786190499*tan(86.4)

h = 2441.60531869599

h = 2441.61 km  is how high the satellite is above the ground.

Return to triangle ACD. We'll use the cosine rule to determine the length of the hypotenuse AC

cos(angle) = adjacent/hypotenuse

cos(A) = AD/AC

cos(86.4) = x/AC

cos(86.4) = 153.612786190499/AC

AC*cos(86.4) = 153.612786190499

AC = 153.612786190499/cos(86.4)

AC = 2446.43279498247

AC = 2446.43 km is the distance from the satellite to station A.

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----------------------------------------------------------------------------

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