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Kaylis [27]
3 years ago
11

Describe the interval(s) on which the function is continuous. (enter your answer using interval notation.)f(x) = xx + 6

Mathematics
1 answer:
Verdich [7]3 years ago
6 0
F(x)=xx+6 = 1(x) =xx+7
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Is 5/6 greater than 24/30
Harrizon [31]

Answer:

Yes

Step-by-step explanation:

The fraction 4/5 is equivalent to 24/30 (To get this I multiplied both the numerator and denominator by 6). The fraction 5/6 is equivalent to 25/30 (To get this I multiplied both the numerator and denominator by 5).... Since 25 is greater than 24, 25/30 (or originally 5/6) is the larger fraction

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PLEASE HELP IM STUCK
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Step-by-step explanation:

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Solve 6 + 5 √ 2 4 9 − 2 x = 7
NISA [10]

6+5\sqrt{249}-2x=7 \\-2x=7-6-5\sqrt{249} \\-2x\approx-77.9 \\x\approx\frac{-77.9}{2}\approx38.95

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Rudiy27

Answer:

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3 0
2 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.

This is the pvalue of Z when X = 85 subtracted by the pvalue of Z when X = 70. So

X = 85

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

Z = \frac{85 - 74}{2.7}

Z = 4.07

Z = 4.07 has a pvalue of 1

X = 70

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

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.

7 0
2 years ago
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