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Rashid [163]
2 years ago
15

How do you solve a-3.5=14.9

Mathematics
2 answers:
just olya [345]2 years ago
4 0
Add 3.5 to 14.9------------
Gennadij [26K]2 years ago
4 0
You would add 3.5 to both sides in order to cancel out the -3.5 on the left side of the equation. This would give you the answer of a = 18.4.
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80 as a product of prime factors
laila [671]
80
8 x 10
2 x 4 x 5 x 2
2 x 2 x 2 x 2 x 5 <===
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2 years ago
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What is 4and1/2 subtract 1/4
umka21 [38]

4 1/2=4 2/4

4 2/4-1/4=4 1/4 or 4.25

8 0
3 years ago
(02.01)
malfutka [58]

Answer:

the answer is Any Real Number

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Step-by-step explanation:

7 0
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(12 - 5)2 + (83 - 9x6)
Lesechka [4]

Answer:

78

Step-by-step explanation:

Solve by PEMDAS.

Parentheses first.

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(7)^2+(29)

Exponent next,

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add them up and you get 78.

Hope this helps! Hit the crown if you want.

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3 years ago
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Find the Z-scores that separate the middle 61% of the distribution from the area in the tails of the standard normal distributio
professor190 [17]

Answer:

Z-scores between -0.86 and 0.86 separate the middle 61% of the distribution from the area in the tails of the standard normal distribution

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Middle 61%

Between the 50 - (61/2) = 19.5th percentile and the 50 + (61/2) = 80.5th percentile.

19.5th percentile.

Z with a pvalue of 0.195. So Z = -0.86

80.5th percentile.

Z with a pvalue of 0.805. So Z = 0.86.

Z-scores between -0.86 and 0.86 separate the middle 61% of the distribution from the area in the tails of the standard normal distribution

8 0
2 years ago
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