Annual income of A is 10% more than of B whereas income of B is 20% more than that of C. If monthly income of C is $ 2000 then w

Answer:

The domain includes all real numbers less than or equal to zero.

Step-by-step explanation:

Given the function:

The radicand (expression under the square root) must be non-negative:

Solving for x:

Multiplying both sides by -1 reverses inequality:

Thus, the domain is (-∞, 0].

All real numbers up to and including zero.

The function's range corresponds to [0, ∞).

Refer to the included diagram for clearer visualization.

All real numbers from zero upward.

Evaluating each statement:

Case 1) The domain is all real numbers.

This is incorrect because the domain is constrained to values less than or equal to zero.

Case 2) The range is all real numbers.

This is also false since the output values are zero or positive only.

Case 3) The domain consists of all real numbers less than or equal to zero.

This is correct.

Case 4) The range is all real numbers less than or equal to zero.

This is false as the range is from zero upwards.

We can apply the formula to tackle this problem, utilizing the following variables: R represents the rate, W signifies the workers/things, T stands for time, and J indicates the jobs/things. In the scenario where 10 printers print 10 pages within 10 seconds, we identify W as 10 printers, T as 10 seconds, and J as 10 pages. Thus, we can deduce how many seconds it will take for 50 printers to produce 50 pages. Based on the previous findings, we can determine T. Ultimately, it will still take 10 seconds for 50 printers to print 50 pages.

Response:

(a) The linear depreciation function relating to the car's value y after a few years t is given by:

y = 25,425 - 2,739 × t

(b) The projected worth of the car in 7 years amounts to $6,252.

(c) The depreciation is at a rate of $2,739 yearly.

Step-by-step explanation:

The manufacturer's suggested retail price (MSRP) for the vehicle is $25,425.

Its expected valuation in 5 years is $11,730.

(a) The linear depreciation formula is as follows:

Where;

Asset Cost = $25,425

Salvage Value = $11,730

Asset Lifespan = 5 years

From this, we obtain;

Thus, the linear depreciation function can be expressed as:

y - 25,425 = -2,739×(t - 0)

y = -2,739·t + 0 + 25,425 = 25,425 -2,739·t

y = 25,425 - 2,739 × t

Where;

y = The estimated worth of the car after a specified duration

t = The years calculated for depreciation

(b) The value of the vehicle in 7 years can be calculated as below;

For t = 7, we find;

y = 25,425 -2,739·t = y = 25,425 -2,739 × 7 = $6,252

The expected car value after 7 years is $6,252.

(c)

The car depreciates at a yearly rate of $2,739.

Answer: The right answer is (A).

Step-by-step explanation:

Let X be the daily waste amount produced. X is distributed in a way that is close to Normal with an average of 348.5 pounds and a standard deviation of 38.2 pounds.

P(X > 350) = Normalcdf(lower: 350, upper: 1000, mean: 348.5, SD: 38.2) = 0.484.

Answer:

7 years

Step-by-step explanation:

Since a graph isn’t provided, I will indicate how many years are involved, and you should be able to deduce it from this. First, let’s [use a calculator] to find 2% of 250, which equals 5. Adding 5 to the total only slightly impacts 2% of the original amount, changing it by 0.1 or 5/50. We can continue adding 5 until we exceed 282, which occurs after 7 increments, calculated as (285 minus 250) divided by 5. Thus, the answer is indeed 7 years. A quick calculation will confirm this.