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Sphinxa [80]
3 years ago
8

If logx= -2, what is x?

Mathematics
1 answer:
horsena [70]3 years ago
7 0

Answer:

Answer C: 0.01

Step-by-step explanation:

As log x = -2 is equal to 10 to the power -2

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18. During a sale, cameras were discounted 30% from their original price of $295.99. How much was each camera during the sale to
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Answer:

207.19 is the cost on sale

Step-by-step explanation:

If there is a discount of 30%, we still have to pay 100-30 = 70%

295.99 * 70%

295.99 * .70

207.193

Rounding to the nearest cent

207.19 is the cost on sale

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Answer:

y = -6x + 2

Step-by-step explanation:

Slope intercept form: y = mx + b

m is the slope

b is the y-intercept

The line pictured goes down 6 for each 1 it moves to the right, slope is rise over run, in this case, -6/1 or just -6

The y-intercept is the point that the function crosses the y-axis, in this case 2

Plug those in to get:

y = -6x + 2

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14 times 16=84+140=224
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What is the exact value of sin 48 degrees?
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.7431448255...

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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

c)

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

d)

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

<u>For Bus:</u>

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

= \dfrac{6}{12} = 0.5

<u>For motorcycle:</u>

= \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
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