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2 years ago

What is 58% of 18 can you please help and explian how u got it

1 answer:

5 0

To obtain 58 percent of 18 multiply 18 by 0.58, because 58 percent is equal to 58/100. That said, your result rounded to two decimals is 10.44.

This answers what is 58% of 18. If you have a currency, such as in the case of 18 dollars, proceed as outlined in the example right below.

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the nth term of a sequence is given by an²+bn where a and b are integers the 2nd term of the sequence is 2 4th term of the seque

stealth61 [152]

Answer:

Step-by-step explanation:

Tn = an²+bn

T2 = a2²+2b

2 = 4a +2b

Multiply all through by 2

4 = 8a +4b

4b = 4-8a

T4 = a4² + 4b

20 = 16a +4b

4b = 20-16a

4-8a = 20-16a

16a -8a = 20-4

8a = 16

a = 16/8 = 2

4b = 4-8a = 4-8(2) = 4-16 = -12

b = -12/4 = -3

T6 = a6² + 6b

T6 = 2(36) + 6(-3) = 72 -18 = 54

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2 years ago

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Point A is at (3, 2.5) and point B is at (3, 1.4) on a coordinate grid. The distance between the two points is ____. Input numbe

Brums [2.3K]

The Answer Is , 2.5-1.4 = 1.1

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3 years ago

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4- A manufacturing process produces items whose weights are normally distributed. It is known that 22.57% of all the items produ

galben [10]

Answer:

$\\ \mu = 118\;grams\;and\;\sigma=30\;grams$

Step-by-step explanation:

We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find $\\ \mu\;and\;\sigma$ .

<h3>First Case: items from 100 grams to the mean</h3>

For finding probabilities that corresponds to z-scores, we are going to use here a <u>Standard Normal Table </u><u><em>for cumulative probabilities from the mean </em></u><em>(Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) </em>that is, the "probability that a statistic is between 0 (the mean) and Z".

A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is <em>below the mean</em> ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.

Then

$\\ z = -0.6\;and\;z = \frac{x-\mu}{\sigma}$

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

<h3>Second Case: items from the mean up to 190 grams</h3>

We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.

Then

$\\ z =2.4\;and\; z = \frac{x-\mu}{\sigma}$