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sveta [45]
3 years ago
15

108 is the product of craig's age and 9

Mathematics
1 answer:
damaskus [11]3 years ago
7 0

<em>Craig is 12 years old.</em>

<h2>Explanation:</h2>

____________________________________

<em>I think your complete question is:</em>

<em />

<em>Translate this sentence into a equation. 108 is the product of Craig's age and 9. Use the variable c to represent Chau's age.</em>

<em>____________________________________</em>

<em />

So let's solve this step by step:

STEP 1. 108.

So we write this number alone:

108

STEP 2. ...is the product of Craig's age and 9

So we need to do a multiplication:

Craig's \ age:C \\ \\ C\times9

STEP 4. 108 is the product of Craig's age and 9

The complete statement is:

108=C\times 9 \\ \\ \\ Isolating \ C: \\ \\ C=\frac{108}{9} \\ \\ \boxed{C=12}

Finally, <em>Craig is 12 years old.</em>

<h2>Learn more:</h2>

Parametric equations: brainly.com/question/10022596

#LearnWithBrainly

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Anybody able to help out ????
kodGreya [7K]

Answer:

z=110

a=69

b=47

c=116

d=93

e=86

Step-by-step explanation:

6 0
3 years ago
Sid goes to the store and buys a shirt for $65. The store has sales tax rate of 5.4%. How much tax will Sid have to pay?
mojhsa [17]

Answer:

$3.51

Step-by-step explanation:

We change 5.4% into a decimal which is .054, we then multiply 65 by .054

$65  x  .054 = $3.51

3 0
3 years ago
After the 4th half life how much of the sample is left if the mass is 100g.
vazorg [7]

Answer:

\huge\boxed{\sf 6.25 \ g}

Step-by-step explanation:

After every Half life , Half of the mass is left.

After 1st Half life = 100 g / 2 = 50 g

After 2nd Half life = 50 / 2 = 25 g

After 3rd Half life = 25 / 2 = 12.5 g

After 4th Half life = 12.5 / 2 = 6.25 g

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
6 0
3 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

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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