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3 years ago

What is 35.6 × 10−4 in standard form?

Mathematics

1 answer:

Fittoniya [83]3 years ago

8 0

$35.6\times10^{-4}=35.6\times0.0001=0.00356$

$a^{-n}=\dfrac{1}{a^n}\\\\10^{-4}=\dfrac{1}{10^4}=\dfrac{1}{10000}=0.0001$

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500m:2km in it's lowest form

blagie [28]

500m:2km (change km to m)

500m:2000m (divide by 100)

5m:20m (divide by 5)

1m:4m

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3 years ago

What is the answer to x²+12x+32

maxonik [38]

Answer:

(x+4)(x+8)

Step-by-step explanation:

Factor x²+12x+32 using the AC method.

6 0

2 years ago

Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars

dem82 [27]

Answer:

Step-by-step explanation:

From the information given:

the rate of the cars = $\dfrac{1}{5} \ car / min = 0.2 \ car /min$

the rate of the buses = $\dfrac{1}{10} \ bus / min = 0.1 \ bus /min$

the rate of motorcycle = $\dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min$

The probability of any event at a given time t can be expressed as:

$P(event \ (x) \ in \ time \ (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}$

∴

(a)

$P(2 \ car \ in \ 20 \ min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}$

$P(2 \ car \ in \ 20 \ min) =0.1465$

$P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}$

$P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422$

$P ( 0 \ buses \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}$

$P ( 0 \ buses \ in \ 20 \ min) = 0.1353$

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

$= 0.3333 \times \dfrac{1}{4}$

= 0.0833

∴

$P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}$

P(zero exact change in 10 minutes) = 0.4347

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

$P( 4 \ motorcyles \ in \ 45 \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}$

$P( 4 \ motorcyles \ in \ 45 \ minutes) =0.0469$

Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

For Bus:

P(no other vehicle other vehicle arrives within 5 minutes is)

= $\dfrac{6}{12} = 0.5$

For motorcycle:

$= \dfrac{2 }{12} = \dfrac{1 }{6}$

∴

The required probability = $1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!} \Bigg)$

= 1- 0.5134

= 0.4866

6 0

3 years ago

Susan drove at an average speed of 30 miles per hour for the first 30 miles of a trip and then at an average speed of 60 miles p

maxonik [38]

Answer: option B is the correct answer.

Step-by-step explanation:

Susan drove at an average speed of 30 miles per hour for the first 30 miles of a trip.

Time = distance/speed

This means that time spent by Susan in travelling the first 30 miles would be

Time = 30/30 = 1 hour

He travelled at an average speed of 60 miles per hour for the remaining 30 miles of the trip.

Time = 30/60 = 0.5 hours

Total distance travelled is 30 + 30 = 60 miles.

Total time spent = 1 + 0.5 = 1.5 hours

Speed = distance/time

Average speed = 60/1.5 = 40 miles per hour.

7 0

3 years ago

Factor completely 5x2 − 40.

Varvara68 [4.7K]

The answer will be 5(x^2-8)

7 0

3 years ago