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3 years ago

If I have an equation in which x = pounds, and y = total cost, can I substitute quarts into the equation

Mathematics

1 answer:

Colt1911 [192]3 years ago

8 0

Answer:

Yes, you can

Step-by-step explanation:

As a point of reference, assume the equation is:

$y = 5x$

Where

$y = total\ cost$

and

$x = pounds$

In standard unit of conversion:

$1\ pound = 0.4793\ quart$

So:

$x\ pounds = 0.4793x\ quart$

Substitute 0.4793x for x in $y = 5x$

$y = 5 * 0.4793x$

$y = 2.3965x$

The above equation is the equivalent of $y = 5x$ in quarts

<em>So, irrespective of what the equation is, you can always substitute quarts into the equation.</em>

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Write each degree measure in radians round to the nearest hundredth 25 degree

dusya [7]

Answer:

To convert a degree measure into radians, you multiply it by π/180, π = 3.14 btw, so 25 * π/180 = 25*3.14/180 = 78.5/180 = 0.43633., which is 0.44 to the nearest hundredth.

Step-by-step explanation:

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2 years ago

Solve the equation. (find x)

Irina-Kira [14]

Answer:

Step-by-step explanation:

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3 years ago

The function below can be used to calculate the cost of making a long distance call, f(m), which is based on a $2.50 initial cha

Vera_Pavlovna [14]

F(m) = 2.5 + 0.12m

if Natalie paid $6.82

6.82 = 2.5 + 0.12m
0.12m = 6.82 - 2.5
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m = 4.32 ÷ 0.12
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The call was 36 minutes long.

5 0

3 years ago

10.6.23

Svetach [21]

I = $ 1,200,000.00

Equation:

I = Prt

Calculation:

First, converting R percent to r a decimal

r = R/100 = 3%/100 = 0.03 per year,

then, solving our equation

I = 1000000 × 0.03 × 40 = 1200000

I = $ 1,200,000.00

The simple interest accumulated

on a principal of $ 1,000,000.00

at a rate of 3% per year

for 40 years is $ 1,200,000.00.

3 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

hoa [83]

Answer:

Obese people

$Lower = \mu - 2\sigma = 373- 2(67) = 239$

$Upper = \mu + 2\sigma = 373+ 2(67) = 507$

Lean People

$Lower = \mu - 2\sigma = 526- 2(107) = 312$

$Upper = \mu + 2\sigma = 526+ 2(107) = 740$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, "almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ)".

Solution to the problem

Obese people

Let X the random variable that represent the minutes of a population (obese people), and for this case we know the distribution for X is given by:

$X \sim N(373,67)$

Where $\mu=373$ and $\sigma=67$

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

$Lower = \mu - 2\sigma = 373- 2(67) = 239$

$Upper = \mu + 2\sigma = 373+ 2(67) = 507$

Lean People

Let X the random variable that represent the minutes of a population (lean people), and for this case we know the distribution for X is given by:

$X \sim N(526,107)$

Where $\mu=526$ and $\sigma=107$

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

$Lower = \mu - 2\sigma = 526- 2(107) = 312$

$Upper = \mu + 2\sigma = 526+ 2(107) = 740$

The interval for the lean people is significantly higher than the interval for the obese people.

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3 years ago