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Montano1993 [528]
4 years ago
7

I need help with this question?

Mathematics
2 answers:
Viktor [21]4 years ago
8 0
8 and 18 are the missing values.
madam [21]4 years ago
5 0
I have the answer attached.

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Lubov Fominskaja [6]
The answer is D!!!!!!!!!!!!!
6 0
2 years ago
Find the sum: 28,35,42,49,56,63,70
lianna [129]

Answer:

343

Step-by-step explanation:

Big brain calculator work

28 + 35 + 42 + 49 + 56 + 63 + 70 = 343

5 0
3 years ago
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How do you simplify this 2x+15(9-9x)2x-15(16x-2)-x
valentinak56 [21]

Answer:

-270x^2+31x+30

Step-by-step explanation:

I hope this helps!

4 0
3 years ago
An article suggests that substrate concentration (mg/cm3) of influent to a reactor is normally distributed with μ = 0.50 and σ =
Delvig [45]

Answer:

0.1056 = 10.56% probability that the concentration exceeds 0.60

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 = 0.5, \sigma = 0.08

What is the probability that the concentration exceeds 0.60?

This is 1 subtracted by the pvalue of Z when X = 0.6. So

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

Z = \frac{0.6 - 0.5}{0.08}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

1 - 0.8944 = 0.1056

0.1056 = 10.56% probability that the concentration exceeds 0.60

6 0
3 years ago
What is m A)30<br> B)60<br> C)90<br> D)50
Sergeu [11.5K]

Answer:

A) 30°

Step-by-step explanation:

All triangles angles add always to 180°, so we have two already 75° + 75° = 150° thus the angle is 30°

5 0
2 years ago
Read 2 more answers
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