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gayaneshka [121]
3 years ago
7

For a given input value rrrr, the function hhhh outputs a value qqqq to satisfy the following equation.

Mathematics
1 answer:
trapecia [35]3 years ago
5 0
65fjeiidhdhdsbisjsisbsbsbs
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Cuantos paquetes de 5/8 de kilo harán falta para envasar 125 kilos de harina
bonufazy [111]

Step-by-step explanation:

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4 0
3 years ago
Dear Math Helper,
valentina_108 [34]

Answer:

n = 98, that is, she scored at the 98th percentile.

Step-by-step explanation:

When the distribution is normal, we use 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.

In this question, we have that:

She scored 38, so X = 38

Test scores are normally distributed with a mean of 25 and a standard deviation of 6.4.

This means that \mu = 25, \sigma = 6.4

Find the percentile:

We have to find the pvalue of Z. So

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

Z = \frac{38 - 25}{6.4}

Z = 2.03

Z = 2.03 has a pvalue of 0.98(rounding to two decimal places).

So n = 98, that is, she scored at the 98th percentile.

5 0
3 years ago
Simplify this e5 x e4​
Gre4nikov [31]

Answer:

hope it will help you

Step-by-step explanation:

follow me and like it.

7 0
2 years ago
what is the approximate volume of a sphere? use 3.14 to approximate pi and round your answer to the nearest hundredth if necessa
Strike441 [17]
Is there answer choices? does it give you any numbers for the sphere


3 0
3 years ago
Read 2 more answers
) Find the coordinates of the point A' which is the symmetric point
soldier1979 [14.2K]

Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point  A(3, 2) with respect to the line 2x + y - 12 = 0.​

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

C( \dfrac{x+3}{2}, \dfrac{y+2}{2})

Also, product of slope will be -1 .( Since, they are parallel )

\dfrac{y-2}{x-3} \times -2  = -1\\\\2y - 4 = x - 3\\\\2y - x = 1

x = 2y - 1

So, C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})

Also, C satisfy given line :

2\times ( \dfrac{2y + 2}{2})  + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}

Also,

x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}

Therefore, the symmetric points isA'(\dfrac{31}{5}, \dfrac{18}{5}) .

7 0
2 years ago
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