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Vladimir79 [104]
4 years ago
5

A truck costs 80,000. it depreviates in value 6,000 per year

Mathematics
1 answer:
Lilit [14]4 years ago
7 0

Answer:in one year it will be 74000

Step-by-step explanation:

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20

Step-by-step explanation:

use the cos or sin function to solve

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<h2>cos(∠B) = CB/AB = ED/AD</h2>

Step-by-step explanation:

cosine=\dfrac{adjacent}{hypotenuse}\\\\adjacent=BC\\hypotenuse=AB\\\\\cos B=\dfrac{BC}{AB}\\\\\angle B\cong\angle D,\ \text{therefore}\ \cos B=\cos D.\\\\adjacent=ED\\hypotenuse=AD\\\\\cos B=\cos D=\dfrac{ED}{AD}

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Write an algebraic expression to represent below verbal expression.
mestny [16]

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12(n-3)

Step-by-step explanation:

Let

n -----> the number

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2) The algebraic expression of "the product of 12 and the sum of a number and  negative 3" is equal to multiply by 12 the sum of a number and  negative 3

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Ask Your Teacher The level of nitrogen oxides (NOX) in the exhaust after 50,000 miles or fewer of driving of cars of a particula
Georgia [21]

Answer:

The level is L = 0.084

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 0.08, \sigma = 0.01, n = 36, s = \frac{0.01}{\sqrt{36}} = 0.0017

What is the level L such that the probability that the average NOX level x for the fleet is greater than L is only 0.01?

This is X when Z has a pvalue of 1-0.01 = 0.99. So it is X when Z = 2.325.

Z = \frac{X - \mu}{\sigma}

By the Central limit theorem

Z = \frac{X - \mu}{s}

2.325 = \frac{X - 0.08}{0.0017}

X - 0.08 = 2.325*0.0017

X = 0.084

The level is L = 0.084

4 0
4 years ago
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