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2 years ago

An object is moving at a speed of 9 inches per minute. Express this speed in yards per month.

Mathematics

1 answer:

Whitepunk [10]2 years ago

6 0

Answer:

10,800 yd/mo

Step-by-step explanation:

$\dfrac{9\text{ in}}{\text{min}}=\dfrac{9\text{ in}}{\text{min}}\times\dfrac{1\text{ yd}}{36\text{ in}}\times\dfrac{60\text{ min}}{1\text{ h}}\times\dfrac{24\text{ h}}{1\text{ da}}\times\dfrac{30\text{ da}}{1\text{ mo}}=\dfrac{(9)(60)(24)(30)}{36}\text{ yd/mo}\\\\=\boxed{10,\!800\text{ yd/mo}}$

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What is the product of (-4)(7) and why?

Evgesh-ka [11]

Answer:

The product of (-4)(7) is -28

Step-by-step explanation:

First you will multiply the number together.

This gets you -28.

The reason it is a negative number is because when you multiply a positive and a negative number together, you get a negative number.

Whereas, if it were two negative numbers, (-4)(-7), or two positive numbers,

(4)(7), the answer would have been positive, 28.

3 0

3 years ago

Kendra is considering two different long-distance phone plans. Phone plan A charges a $100 sign-up fee and 3 cents per minute. P

Fed [463]

Answer:

Part 1) The number of minutes in a month must be greater than 50 in order for the plan A to be preferable

Part 2) The number of minutes in a month must be equal to 50 minutes

Step-by-step explanation:

The question is

Part 1) How many minutes would Kendra have to use in a month in order for the plan A to be preferable? Round your answer to the nearest minute

Part 2) Enter the number of minutes where Kendra will pay the same amount for each long distance phone plan

Part 1)

Let

x ---> the number of minutes

we have

Cost Plan A

$3x+100$

Cost Plan B

$5x$

we know that

In order for plan A to be cheaper than plan B, the following inequality must hold true.

cost of plan A < cost of plan B

substitute

$3x+100 < 5x$

solve for x

subtract 3x both sides

$100< 5x-3x\\100$

divide by 2 both sides

$50 < x$

Rewrite

$x> 50\ min$

therefore

The number of minutes in a month must be greater than 50 in order for the plan A to be preferable

Part 2)

Let

x ---> the number of minutes

we have

Cost Plan A

$3x+100$

Cost Plan B

$5x$

we know that

In order for plan A cost the same than plan B, the following equation must hold true.

cost of plan A = cost of plan B

substitute

$3x+100= 5x$

solve for x

$5x-3x=100\\2x=100\\x=50\ min$

therefore

The number of minutes in a month must be equal to 50 minutes

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3 years ago

How do you find how many times larger

masya89 [10]

Answer

whats the problem.

Step-by-step explanation:

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4 years ago

Which phase of inferential statistics is sometimes considered to be the most crucial because errors in this phase are the most d

Alex_Xolod [135]

The phase of inferential statistics which is sometimes considered to be the most crucial because errors in this phase are the most difficult to correct is "data gathering".

<h3>What is inferential statistics?</h3>

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Some characteristics of inferential statistics are-

Inferential statistics compare treatments groups and make conclusions about the greater population of participants using measures from the experiment's sample of subjects.
Inferential statistics aids in the development of explanations for a condition or phenomenon.
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To know more about the inferential statistics, here

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2 years ago

(X+15)+(x+5)+(x+15)+(X+5)

Annette [7]

Answer:

4 + 4 0

Step-by-step explanation:

Simplify

Eliminate redundant parentheses

( + 1 5 ) + ( + 5 ) + ( + 1 5 ) + ( + 5 )

Add the numbers

Combine like terms

Solution

4 + 4 0

6 0

3 years ago