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motikmotik
2 years ago
13

A moving van travels 560 miles to reach a new house. The drivers stop to rest after traveling 75% of the total distance. How man

y miles have the drivers traveled when they stop
Mathematics
1 answer:
dexar [7]2 years ago
5 0

Answer:

the driver stopped when they had traveled 420 miles

Step-by-step explanation:

hope this helps,

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The speed(S) of a car varies partly directly as its mass(M) and partly directly as the quantity (Q) of fuel in it. When the spee
weeeeeb [17]

Answer:

S varies partly directly as M and Q.

S=C.

S=KMQ+C.

For the first one...

speed=80,m=220,Q=30.

80=K20×30+C.

80=600K+C......(I).equation one.

For the second one....

speed=60,m=300,Q=40.

60=K300×40+C.

60=12000K+C.....(ii). equation two.

Minus eqtn(I) from eqtn(ii).

80=600K+C.

- 60=12000K+C.

K=0.01754~0.018.

Substitute K=0.018 into eqtn(I).

80=600K+C

80=600×0.018+C.

80=10.8+C.

C=80-10.8=69.2.

The relation is S=0.018MQ+69.2

when speed is 100 and mass is 250 find the volume.

100=0.018×250×Q+69.2.

100=4.5Q+69.2.

4.5Q=100-69.2

4.5Q=30.8.

Q=30.8/4.5.

Q=6.8~7litres.

6 0
3 years ago
Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
NeX [460]

Answer:

(a) 0.343

(b) 0.657

(c) 0.189

(d) 0.216

(e) 0.353

Step-by-step explanation:

Let P(a vehicle passing the test) = p

                        p = \frac{70}{100} = 0.7  

Let P(a vehicle not passing the test) = q

                         q = 1 - p

                         q = 1 - 0.7 = 0.3

(a) P(all of the next three vehicles inspected pass) = P(ppp)

                           = 0.7 × 0.7 × 0.7

                           = 0.343

(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)

      = (0.3 × 0.7 × 0.7) + (0.3 × 0.3 × 0.7) + (0.7 × 0.3 × 0.7) + (0.7 × 0.3 × 0.3) + (0.7 × 0.7 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.3)

      = 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027

      = 0.657

(c) P(exactly one of the next three inspected passes) = P(pqq or qpq or qqp)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

                 = 0.063 + 0.063 + 0.063

                 = 0.189

(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)

                 = 0.063 + 0.063 + 0.063 + 0.027

                 = 0.216

(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?

P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)

=  (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063

                  = 0.973  

With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,

                         = \frac{P(all\ of\ the\ next\ three\ vehicles\ inspected\ pass)}{P(at\ least\ one\ of\ the\ next\ three\ vehicles\ inspected\ passes)}

                         = \frac{0.343}{0.973}

                         = 0.353

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4 years ago
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