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Katyanochek1 [597]
3 years ago
5

Its line segment please

Mathematics
2 answers:
zavuch27 [327]3 years ago
7 0

Answe]here g is apoint on dj sdj = dg + gj 11x + 25 = dg + 44x + 14 11x - 44x + 25 - 14 = dg - 33x + 11 = dg

11 x + 25=8x+34

x=3

dg= 4×3+17=29

thank you

joja [24]3 years ago
6 0

Answer:

A) 29

Step-by-step explanation:

DG = GJ

GJ = DG = 4x + 17

DJ = DG + GJ

11x + 25 = 4x + 17 + 4x + 17

11x + 25 = 4x + 4x + 17 + 17    {Combine like terms}

11x + 25 = 8x + 34

Subtract 25 from both sides

11x = 8x + 34 - 25

11x = 8x + 9

Subtract 8x from both sides

11x - 8x = 9

3x = 9

Divide both sides by 3

x = 9/3

x = 3

DG =  4x + 17

     = 4*3 + 17

    = 12 + 17

DG = 29

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Answer:

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Step-by-step explanation:

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3 years ago
A study suggested that childrenbetween the ages of 6 and 11 in the US have anaverage weightof 74 lbs. with a standard deviation
Doss [256]

Answer:

The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.

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.

A study suggested that children between the ages of 6 and 11 in the US have an average weightof 74 lbs, with a standard deviation of 2.7 lbs.

This means that \mu = 74, \sigma = 2.7

What proportion of childrenin this age range between 70 lbs and 85 lbs.

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X = 85

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Z = \frac{70 - 74}{2.7}

Z = -1.48

Z = -1.48 has a pvalue of 0.0694

1 - 0.0694 = 0.9306

The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.

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