Increase 400 by 10% and then decrease the answer by 10%

Mathematics

2 answers:

makvit [3.9K]2 years ago

8 0

Answer:80

Assume the number to be 400 for instance,

10% of 400 increased = 400 + ( 10/100 × 400 )

= 400 + 40 = 440

The resulting number decreased by its 10% = 400 - (10/100 × 400)

= 400 - 40 = 360

Now, net decrease = 440 - 360 = 80

erik [133]2 years ago

7 0

Answer:

400.

10% of 400 is 40 (Simply reduce the hundreds place to the tens place)

So now we have 440.

Decrease it by 10% again, and that brings you right back to 400.

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How do you factor quadratics by grouping

Kaylis [27]

Here's an example Its also call the 'ac' method:-

factor 2x^2 - 13x - 15

Multiply first coefficient by last number which is 2*-15 = -30

Now we need 2 factors of -30 which will when added give - 13:-

-15 and 2 looks good. So we can write:-

2x^2 - 13 x - 15

= 2x^2 + 2x - 15x - 15

factor by grouping first 2 and last 2 terms , we get:-

= 2x(x + 1) - 15(x + 1)

the x+1 is common so we have:-

(2x - 15)(x + 1) Answer

Hope this helps.

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yanalaym [24]

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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 are 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 sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .