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3 years ago

- x+5y = 19 X+ 2 y = 16 Solving systems using elimination

Mathematics

2 answers:

Vesna [10]3 years ago

7 0

Hope this helps with the answer ;)

alina1380 [7]3 years ago

6 0

Answer:

Step-by-step explanation:

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What is the quotient of 67,860 ÷ 13? 4,120 4,220 5,120 5,220

IRISSAK [1]

Answer:

If 67860÷13,then quotient =5220Answer

6 0

3 years ago

Examine the diagram of circle C. Points Q, V, and W lie on circle C. Given that m∠VCW=97∘, what is the length of VW⌢?

8090 [49]

Answer:

$\frac{97pi}{18} m$

Step-by-step Explanation:

==>Given:

radius (r) = 10 m

m<VCW = 97°

==>Required:

Length of arc VW

==>Solution:

Formula for length VW is given as 2πr(θ/360)

Using the formula, Arc length = 2πr(θ/360), find the arc length VW in the given circle

Where,

θ = 97°

r = radius of the circle = 10 m

Thus,

Arc length VW = 2*π*10(97/360)

Arc length VW = 20π(97/360)

Arc length VW = π(97/18)

Arc length VW = 97π/18 m

Our answer is,

$\frac{97pi}{18} m$

8 0

4 years ago

The Dow Jones Industrial Average has had a mean gain of 432 pear year with a standard deviation of 722. A random sample of 40 ye

trasher [3.6K]

Answer:

66.98% probability that the mean gain for the sample was between 250 and 500.

Step-by-step explanation:

To solve this problem, it is important to know 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .