Solve for the portion. Round to hundredths when necessary.What is the portion if the base is 1,850 and the rate is 150%?

Answers

Answer 1

Answer:

The portion is $2,475 if the base is 1,850 and the rate is 150%.

What are Arithmetic operations?

Arithmetic operations can also be specified by subtracting, dividing, and multiplying built-in functions. The operator that performs the arithmetic operation is called the arithmetic operator.

* Multiplication operation: Multiplies values on either side of the operator

For example 12*2 = 24

We have to determine the portion if the base is 1,850.

Here rate is 150% or 150/100 = 1.50.

The base = 1,650.

Since the portion is the multiplication of base and rate.

The portion = base × rate.

The portion = 1650 × 1.50.

The portion = $2,475

Therefore, the portion is $2,475 if the base is 1,850 and the rate is 150%.

Answer 2

Answer:

Answer:

portion = base * rate. rate is 150% or 150/100 = 1.50. base = 1,650. portion = 1650*1.50. portion = $2,475

Step-by-step explanation:

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An experimenter is studying the effects of temperture, pressure, and type of catalyst on yield from a certain chemical reaction. Three different temperatures, four different pressures, and five different catalysts are under consideration.(a) If any particular experiental run involves the use of single temperature, pressure, and catalyst, how many experimental runs are possible?
(b) How many experimental runs are there that involve use of the lowest temperature and two lowest pressure?

Answers

Answer:

Step-by-step explanation:

Given that an experimenter is studying the effects of temperture, pressure, and type of catalyst on yield from a certain chemical reaction. Three different temperatures, four different pressures, and five different catalysts are under consideration.

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A car travels 85 miles on 5 gallons of gas. Give the ratio of miles to gallons as a rate in miles per gallon.

Answers

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An attacker at the base of a castle wall 4 m high throws a rock straight up with speed 7.5 m/s from a height of 1.5 mabove the ground.
(a) Will the rock exceed the top of the wall?
(b) If so, what is its speed when it reaches the top of the wall?
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(c) If we don’t want the stone to exceed the top of the wall (we want the top of the wall to be the maximum height)
what the initial speed that the stone must have?

Answers

Answer:

a) Yes, b) $v\approx 2.686\,(m)/(s)$ , c) $v_(o) \approx 7.002\,(m)/(s)$

Step-by-step explanation:

a) The maximum height reached by the rock is:

$\Delta y = (\left(0\,(m)/(s) \right)^(2)-\left(7.5\,(m)/(s) \right)^(2))/(2\cdot \left(-9.807\,(m)/(s^(2)) \right))$

$\Delta y = 2.868\,m$

$y = 1.5\,m + 2.868\,m$

$y = 4.368\,m$

Yes, the rock will exceed the top of the wall.

b) The speed when the rock reaches the top of the wall:

$v = \sqrt{\left(7.5\,(m)/(s) \right)^(2)+2\cdot \left(-9.807\,(m)/(s^(2)) \right)\cdot (4\,m-1.5\,m)}$

$v\approx 2.686\,(m)/(s)$

c) The initial speed required so that stone does not exceed the top of the wall is:

$v_(o) = \sqrt{\left(0\,(m)/(s) \right)^(2)-2\cdot \left(-9.807\,(m)/(s^(2)) \right)\cdot (4\,m-1.5\,m) }$

$v_(o) \approx 7.002\,(m)/(s)$

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Answers

Answer:

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Step-by-step explanation:

Cost to store: $140

Markup: 25%

The selling price= $140 + 25% = $140*1.25= $175

PLZ HELP ASAP CONGRUENCE

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