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3 years ago

Let's find the unknown variables in the following cases. ) (a) If C.P. Rs 150 and profit = Rs 12, find profit percent

Mathematics

2 answers:

vivado [14]3 years ago

6 0

Answer:

solution, cp=Rs150 profit=Rs12 Now, profit percent=profit% of cp or,profit percent=12/100 x Rs150 Therefore,profit percent =18%

Step-by-step explanation:

Romashka [77]3 years ago

3 0

Answer:

Step-by-step explanation:

We know that

Profit% = Profit/CP × 100

=> 12/150 × 100

=> 12/15 × 10

=> 4/5 × 10

=> 8%

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Can any one please help me I really need help please help me thank you

lina2011 [118]

Answer:

Jack has to sell the old computers worth $8000

Step-by-step explanation:

Jack is paid for selling used computers = $15 per hour plus 5% commission

Let the revenue generated from selling the computers by him = $x

He will be paid for 40 hours = $(40×15) = $600

Commission earned by selling x computers = $0.05x

Total amount earned by him in 40 hours week = 600 + 0.05x

If this amount is $1000,

$1000 = $(600 + 0.05x)

0.05x = 1000 - 600

0.05x = 400

x = $\frac{400}{0.05}$

x = $8000

Therefore, Jack has to sell the old computers worth $8000.

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3 years ago

Suppose that an airline quotes a flight time of 128 minutes between two cities. Furthermore, suppose that historical flight reco

ANTONII [103]

Answer:

(a) The probability density function of X is:

$f_{X}(x)=\frac{1}{b-a};\ a$

(b) The value of P (129 ≤ X ≤ 146) is 0.3462.

(c) The probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.

Step-by-step explanation:

The random variable X is defined as the flight time between the two cities.

Since the random variable X denotes time interval, the random variable X is continuous.

(a)

The random variable X is Uniformly distributed with parameters a = 10 minutes and b = 154 minutes.

The probability density function of X is:

$f_{X}(x)=\frac{1}{b-a};\ a$

(b)

Compute the value of P (129 ≤ X ≤ 146) as follows:

Apply continuity correction:

P (129 ≤ X ≤ 146) = P (129 - 0.50 < X < 146 + 0.50)

= P (128.50 < X < 146.50)

$=\int\limits^{146.50}_{128.50} {\frac{1}{154-102}} \, dx$

$=\frac{1}{52}\times \int\limits^{146.50}_{128.50} {1} \, dx$

$=\frac{1}{52}\times (146.50-128.50)$

$=0.3462$

Thus, the value of P (129 ≤ X ≤ 146) is 0.3462.

(c)

It is provided that a randomly selected flight between the two cities will be at least 3 minutes late, i.e. X ≥ 128 + 3 = 131.

Compute the value of P (X ≥ 131) as follows:

Apply continuity correction:

P (X ≥ 131) = P (X > 131 + 0.50)

= P (X > 131.50)

$=\int\limits^{154}_{131.50} {\frac{1}{154-102}} \, dx$

$=\frac{1}{52}\times \int\limits^{154}_{131.50} {1} \, dx$

$=\frac{1}{52}\times (154-131.50)$

$=0.4327$

Thus, the probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.

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zzz [600]

Answer:

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Step-by-step explanation:

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3 years ago

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RSB [31]

Hey there!

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4 years ago

What series of transformations map triangle △ABC onto △EDF to prove that ABC≅EDF ?

Nonamiya [84]

Hello!

The correct answer is A translation 3 units up then a reflection across the x-axis.

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3 years ago

Read 2 more answers

Let's find the unknown variables in the following cases. ) (a) If C.P. Rs 150 and profit = Rs 12, find profit percent​

Let's find the unknown variables in the following cases. ) (a) If C.P. Rs 150 and profit = Rs 12, find profit percent