The graph of y = 2x2 + 10x + 12 is shown. If the

Shishir bought 4000 orange at 70 paisa each. But 400 of them were rotten. He sold 2000 oranges at 90 paisa each.If he plans to m

Ipatiy [6.2K]

Answer:

He needs to sell the rest of the oranges at 75 paisa each.

Step-by-step explanation:

Consider the given information that, Shishir bought 4000 orange at 70 paisa each.

Note: 1 rupees = 100 paisa

Thus, 70 paisa = 70/100 rupees = 0.70 rupees

Therefore, the cost price of 4000 oranges is:

4000×0.70 rupees = 2800 rupees

The selling price of 2000 oranges is:

2000×0.90 rupees = 1800 rupees

The number of oranges now Shishir have:

4000 - 2000 - 400 = 1600

He wants to make a profit of RS 200. Thus the selling price of 4000 oranges should be:

2800 rupees + 200 rupees = 3000 rupees

He earned 1800 rupees by selling 2000 oranges at 90 paisa. So, the remaining amount that he needs to make with 1600 oranges is:

3000 rupees - 1800 rupees = 1200 rupees

Therefore, the cost of one orange is:

1600 oranges = 1200 rupees

1 orange = 1200/1600 rupees

1 orange = 0.75 rupees

Hence, he needs to sell the rest of the oranges at 75 paisa each.

4 0

2 years ago

FWML is a parallelogram. Find the values of x and y. Solve for the value of z, if z=x−y.

Ganezh [65]

Answer:

x = 5, y = 8, z = -3

Step-by-step explanation:

Opposite sides of a parallelogram are congruent so to find x:

x + 7 = 3x - 3

-2x = -10

x = 5

To find y:

y + 2 = 2y - 6

-y = -8

y = 8

Therefore, z = x - y = 5 - 8 = -3.

4 0

2 years ago

A company buys paper clips at the rate of $7.50 per box for the first 10 boxes, $5.50 per box for the next 10 boxes, and $3.50 p

BlackZzzverrR [31]

Answer:

46 boxes

Step-by-step explanation:

First 10 boxes can be bought for $7.5*10=75$
Next 10 boxes can be bought for $5.5*10=55$

So for a total of 20 boxes, the cost is $75+55=130$

We have remaining $221-130=91$

Each additional box costs 3.5, so for 91 dollars, we can get:

$\frac{91}{3.5}=26$ boxes

So, total $10+10+26=46$ boxes can be bought for $221.00

5 0

2 years ago

Part A: During what interval(s) of the domain is the water balloon's height increasing?

Advocard [28]

Answer:

The answer is below

Step-by-step explanation:

The linear model represents the height, f(x), of a water balloon thrown off the roof of a building over time, x, measured in seconds: A linear model with ordered pairs at 0, 60 and 2, 75 and 4, 75 and 6, 40 and 8, 20 and 10, 0 and 12, 0 and 14, 0. The x axis is labeled Time in seconds, and the y axis is labeled Height in feet. Part A: During what interval(s) of the domain is the water balloon's height increasing? (2 points) Part B: During what interval(s) of the domain is the water balloon's height staying the same? (2 points) Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest? Use complete sentences to support your answer. (3 points) Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 16 seconds.

Answer:

Part A: During what interval(s) of the domain is the water balloon's height increasing?

Between 0 and 2 seconds, the height of the balloon increases from 60 feet to 75 feet

Part B: During what interval(s) of the domain is the water balloon's height staying the same?

Between 2 and 4 seconds, the height remains the same at 75 feet. Also from 10 seconds the height of the balloon is at 0 feet

Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest?

Between 4 and 6 seconds, the height of the balloon decreases from 75 feet to 40 feet (i.e. -17.5 ft/s)

Between 6 and 8 seconds, the height of the balloon decreases from 40 feet to 20 feet (i.e. -10 ft/s)

Between 8 and 10 seconds, the height of the balloon decreases from 20 feet to 0 feet (i.e. -10 ft/s)

Hence it decreases fastest from 4 to 6 seconds

Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 16 seconds

From 10 seconds, the balloon is at the ground, so it remains at the ground (0 feet) even at 16 seconds

6 0

2 years ago

If the size of the sample to be used in a particular test of attributes has not been determined by utilizing statistical concept

EleoNora [17]

Answer:

C) The auditor may or may not achieve the desired risk of assessing control risk too low.

Step-by-step explanation:

In a concept of risk sampling, if the sample size is chosen randomly in accordance with random selection procedures, the auditor may or may not achieve the desired risk of assessing risk too low. In other words the auditor may or may not achieve desired precision. This is because a samole chosen randomly may not represent the true population.

This depends largely on the sample size. If the sample size selected is too small, the allowance for sampling risk will be larger than what is required because it will lead to a large standard error of the mean

5 0

2 years ago