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Vsevolod [243]
4 years ago
15

PLS HELP 50PTS SOLVE FOR X

Mathematics
2 answers:
GenaCL600 [577]4 years ago
8 0

Answer:

x= 13


Step-by-step explanation:


denis23 [38]4 years ago
8 0

Answer:

x is 13

Step-by-step explanation:

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For which values of X and Y is line P parallel to line Q
zubka84 [21]

Answer:

3rd option x=3, y=5

Step-by-step explanation:


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3 years ago
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See picture for answer.

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4 years ago
There are three modes of transporting material from Ontario to Florida, namely, by land, sea, or air. Also land transportation m
Semenov [28]

Answer:

0.057

0.6140

0.3158

0.0701

Step-by-step explanation:

Given that:

Let :

P(L) = Number transported by land = half = 50% = 0.5

P(S) = number transported by sea = 30% = 0.3

P(A) = Number transported by air = (100 - (50 + 30))% = 20% = 0.2

P(H) = highway transport = 40% of land transport = 0.4 * 0.5 = 0.2

P(R) = Rail shipment =(100- 40)% = 60% of land transport = 0.6 * 0.5 = 0.3

Percentage of damaged cargo :

Let probability of damage = P(d)

P(d | H) = 0.1

P(d | R) = 0.05

P(d | S) = 0.06

P(d | A) = 0.02

1) What percentage of all cargoes may be expected to be damaged

[P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]

(0.1*0.2) + (0.05*0.3) + (0.06*0.3) + (0.02*0.2) = 0.057

(2) If a damaged cargo is received, what is the probability that it was shipped by ;

land?

([P(d | H)*p(H)] + [P(d | R)*p(R)]) / [P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]

((0.1*0.2) + (0.05*0.3)) / (0.1*0.2) + (0.05*0.3) + (0.06*0.3) + (0.02*0.2)

= 0.035 / 0.057

= 0.6140

By sea?

[P(d | S)*p(S)] / [P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]

(0.06 * 0.3) / 0.057

= 0.3158

By air?

[P(d | A)*p(A)] / [P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]

(0.02 * 0.2) / 0.057

= 0.0701

7 0
3 years ago
Jamie needs to determine the distance across a river. She selects a rock, R, on the opposite river bank. She finds a tree, T, on
Setler [38]

Since RT is perpendicular to the river, you have that segment CT is perpendicular to RT and, thus, CT⊥RB, because point B lies in a straight line with points R and B. Angles TCR and TCB are congruent (she walks from C in a direction that makes the same angle as TCR with the river bank on her side of the river).

This means that triangle CRB is isosceles triangle with base BR and height CT, because height CT is also a bisector of angle BCR.

The height CT divides triangle BCR into two right triangles BCT and RCT, where:

  • CT is common leg;
  • ∠TCR≅∠TCB,

then by LA Theorem these two triangles are congruent (ΔBCT≅ΔRCT).

Congruent triangles have congruent corresponding sides:

  • BC≅RC;
  • CT≅CT;
  • BT≅TR.

If BT turns out to be 75 yards, then TR=75 yards.

By the Pythagorean theorem,

BC^2=CT^2+BT^2,\\ \\BC^2=50^2+75^2=2500+5625=8125,\\ \\BC=\sqrt{8125}=25\sqrt{13}.

If BC=25\sqrt{13}, then distance from the spot to the rock is  CR=25\sqrt{13}.

Answer: RT=75 yards, CR=25\sqrt{13} yards.

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3 years ago
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zimovet [89]

Answer:

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