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2 years ago

70 thousand written in standard form or word form explain

Mathematics

2 answers:

trapecia [35]2 years ago

6 0

Word Form: Just write out the number.

Seventy Thousand

Standard Form (To how I was Taught): 70,000 (If you had other digits you would add them in order [Example: 543 = 500+40+3])

dangina [55]2 years ago

4 0

To solve, you must know that thousand = 1000. Multiply the two numbers together

70 thousand = 70 x 1000 = 70,000

70,000 is your standard form.

seventy-thousand is your word form

hope this helps

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Let
x-------------> <span>the price of each cup
</span>y-------------> the price of each plate

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I substitute 2 in 1
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3 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

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Answer:

(a) 0.343

(b) 0.657

(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

= (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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