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Allisa [31]
2 years ago
15

What is the value of the number in the thousandths place 7.819

Mathematics
2 answers:
MrMuchimi2 years ago
8 0

Answer:

the value of the number in the thousandths place is 9

Step-by-step explanation:

No calculator!

kvasek [131]2 years ago
5 0
9 is in the thousandths place
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Alika [10]

Answer:

105.3

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Consider the equation 43 - 5 (2x + 1) = ax + b.
Trava [24]

Answer:

0

Step-by-step explanation:

4 0
2 years ago
A set of exam scores is normally distributed and has a mean of 80.2 and a standard deviation of 11. What is the probability that
ZanzabumX [31]

Answer:

93.32% probability that a randomly selected score will be greater than 63.7.

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 problem, we have that:

\mu = 80.2, \sigma = 11

What is the probability that a randomly selected score will be greater than 63.7.

This is 1 subtracted by the pvalue of Z when X = 63.7. So

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

Z = \frac{63.7 - 80.2}{11}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

1 - 0.0668 = 0.9332

93.32% probability that a randomly selected score will be greater than 63.7.

5 0
3 years ago
How is the series 9 + 13+ 17+ ... + 149 represented in summation notation?
VLD [36.1K]

Notice that

13 - 9 = 4

17 - 13 = 4

so it's likely that each pair of consecutive terms in the sum differ by 4. This means the last term, 149, is equal to 9 plus some multiple of 4 :

149 = 9 + 4k

140 = 4k

k = 140/4

k = 35

This tells you there are 35 + 1 = 36 terms in the sum (since the first term is 9 plus 0 times 4, and the last term is 9 plus 35 times 4). Among the given options, only the first choice contains the same amount of terms.

Put another way, we have

\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k=0}^{35} (9 + 4k)

but if we make the sum start at k = 1, we need to replace every instance of k with k - 1, and accordingly adjust the upper limit in the sum.

\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k-1=0}^{35+1} (9 + 4(k-1))

\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k=1}^{36} (5 + 4k)

7 0
2 years ago
Math 8
Airida [17]
Yes I’ll send u the answers
8 0
3 years ago
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