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djyliett [7]
3 years ago
9

What number round to 200.000 when rounded to the nearest hundred thousand? Choose all that apply

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
7 0

Answer:

Is there multiple choice to this question?

anyway the answer maybe is: 0

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Simplify 5a+6a2+7a-4a2-9a
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Answer:

2a to the second power + 3a

Step-by-step explanation:

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A football team gained 5 yards, lost 8 yards, lost another 2 yards, and then gained 45 yards. What were the total yards gained o
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yes you would be correct if they started with zero yards at the beginning

Step-by-step explanation:

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3 years ago
In step 7, you listed three elements commonly listed in charts. what are all of those options? check all that apply. data points
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Data points
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7 0
3 years ago
Is - 11>-16 a true pr false statement
Elina [12.6K]

Answer:

True

Step-by-step explanation:

-11 is greater than -16

6 0
3 years ago
Test scores are normally distributed with a mean of 500. Convert the given score to a z-score, using the given standard deviatio
Bond [772]

Answer:

The percentage of students who scored below 620 is 93.32%.

Step-by-step explanation:

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.

In this question, we have that:

\mu = 500, \sigma = 80

Percentage of students who scored below 620:

This is the pvalue of Z when X = 620. So

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

Z = \frac{620 - 500}{80}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

The percentage of students who scored below 620 is 93.32%.

3 0
3 years ago
Read 2 more answers
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