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e-lub [12.9K]
3 years ago
11

What is a main disadvantage of leasing a vehicle compared to buying a vehicle?

Mathematics
1 answer:
dimaraw [331]3 years ago
6 0
There’s no cut off with a lease, so buying a vehicle has a set price that you pay until you reach, where as leasing a vehicle means you make the same set payment for the time that you have the vehicle. So, let’s say the price of a vehicle is $15,000 or you can lease the same vehicle for $200 a month. If you have the car for 10 years... (120 months)... you have payed $15000 for buying the car but if you lease the car you would have payed $24,000
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100 points and brainliest<br> What Is the purpose of an amino acid codon chart?
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➷ The purpose of having these is so that we can identify the amino acid that corresponds to DNA and RNA to produce a protein. They specify the order of amino acids in producing the protein.

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8 0
2 years ago
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What the length of a diagonal of a square with perimeter of 16
Vladimir79 [104]
A- the length of the side of a square
P = 16

P = 4a → 4a = 16  |:4 → a = 4

The diagonal of the square: d = a√2

therefore d = 4√2


4 0
3 years ago
What is 40/24 simplified
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8 0
3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

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