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love history [14]
1 year ago
5

In the point (3,-2), what is the x-value? Which quadrant is this point located?

Mathematics
1 answer:
natali 33 [55]1 year ago
5 0

Answer:

Step-by-step explanation:

The x-value is 3. Coordinate pairs are labeled as (x, y). In this case, the 3 is in the x position.

The point is located in the 4th quadrant. Quadrant labels move counter-clockwise, from 1 to 4 as shown below. In Quadrant 4, the x-value will always be positive, and the y-value will always be negative. This point has a positive x-value and negative y-value, so it is in this quadrant.

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For tax purposes, a car rental company assumes each car in their fleet depreciates by 6% per year. If the initial value of a car
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The value of a car after 10 years be $20898.27 if the car rental company assumes each car in their fleet depreciates by 6% per year option (a) $20898.27 is correct.

<h3>What is an exponential function?</h3>

It is defined as the function that rapidly increases and the value of the exponential function is always a positive. It denotes with exponent \rm y = a^x

where a is a constant and a>1

We can solve this problem by exponential function:

The word  depreciate means the price is decreasing.

We can find the value be when the car is 10 years old:

p = 38800(1 - 0.06)¹⁰

p = 38800(0.94)¹⁰

p = 20898.266 ≈ $20898.27

Thus, the value of a car after 10 years be $20898.27 if the car rental company assumes each car in their fleet depreciates by 6% per year option (a) $20898.27 is correct.

Learn more about the exponential function here:

brainly.com/question/11487261

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6 0
2 years ago
Miss Florez buys 1 pound of apples how many ounces of apple does she buy
hram777 [196]
1 pound is 16 ounces so I hope this helps

16 ounces of apples
8 0
4 years ago
As part of his retirement strategy, John plans to invest $210,000 in two different funds. He projects that the moderately high r
Maslowich

Answer:

Therefore he invested $60,000 at 9% per year and $(210,000-60,000)=$150,000 at 4% per year.

Step-by-step explanation:

Given John plans to invest $210,000 in two different funds. He projects that the moderately high risk investments should return, overtime 9% per year,while low risk investments should return about 4% per year.

He wants a supplemental income of$11,400 a year.

Let , he invested $x at 9% per year and $(210,000-x) at 4% per year.

interest=\frac{prt}{100}   p = principle , r = rate of interest and t = time

The interest earns at 9% per year= \frac{x\times 9\times 1}{100}

The interest earns at 4% per year=\frac{(210,00-x)\times 4 \times 1}{100}

According to the problem,

\frac{x\times 9\times 1}{100}+\frac{(210,00-x)\times 4 \times 1}{100}= 11400

\Leftrightarrow 9x+840000-4x=11400 \times 100

\Leftrightarrow 5x=1140000-840000

\Leftrightarrow x=\frac{300000}{5}

\Leftrightarrow x=60,000

Therefore he invested $60,000 at 9% per year and $(210,000-60,000)=$150,000 at 4% per year.

7 0
3 years ago
Susan tossed a die 720 times. Which of the following would be a good estimate of the number of times she got the number 3 on the
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120 out of 720. Hope I heped.
8 0
3 years ago
Suppose you know the length of a confidence interval of a population mean is 8.4 and the sample mean (x bar) is 10. Find the ​ma
SOVA2 [1]

Answer:

ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

The lenght of the interval correspond to:

8.4 = 2ME

ME= \frac{8.4}{2}= 4.2

And since we know the margin of error we can find the limits for the confidence interval:

Lower = 10 -4.2=5.8

Upper = 10 +4.2=14.2

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

\bar X=10 represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n represent the sample size  

Solution to the problem

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma)

The sample mean \bar X is distributed on this way:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})  

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

The margin of error is given by:

ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

The lenght of the interval correspond to:

8.4 = 2ME

ME= \frac{8.4}{2}= 4.2

And since we know the margin of error we can find the limits for the confidence interval:

Lower = 10 -4.2=5.8

Upper = 10 +4.2=14.2

4 0
3 years ago
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