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Lana71 [14]
3 years ago
11

Using distributive property what is 2*(9+1/2)

Mathematics
2 answers:
Alex17521 [72]3 years ago
8 0

Answer:

the answer is 19

Step-by-step explanation:

dmitriy555 [2]3 years ago
3 0

Answer:19

Step-by-step explanation:

2(9+1/2)

2(9)=2(1/2)

18+1

19

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2.6 = 2.600
So, 2.6 is not bigger than 2.661
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If 3 + root8 by 3-root8 + 3-root8 by 3+root8 = a+broot2 find a and b
s2008m [1.1K]

If \frac{3+\sqrt{8} }{3-\sqrt{8} } +\frac{3-\sqrt{8} }{3+\sqrt{8} }=a+b\sqrt{2}, the value of a and b is given as a = 34 and b = 0

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables or numbers.

\frac{3+\sqrt{8} }{3-\sqrt{8} } +\frac{3-\sqrt{8} }{3+\sqrt{8} }\\ \\\frac{9+6\sqrt{8}+8+9-6\sqrt{8}+8  }{(3-\sqrt{8} )(3+\sqrt{8} )} \\\\\frac{34}{9-8} =34+0\sqrt{2}

If \frac{3+\sqrt{8} }{3-\sqrt{8} } +\frac{3-\sqrt{8} }{3+\sqrt{8} }=a+b\sqrt{2}, the value of a and b is given as a = 34 and b = 0

Find out more on equation at: brainly.com/question/2972832

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2 years ago
12÷4×32+(4−2)5
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Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all
posledela

This question is Incomplete

Complete Question

Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all have the same body size in their adult phase, which made it easy to measure speeds in units of body lengths per second (bl/s). The researchers found that, when traffic is light and not congested, ant speeds vary roughly Normally, with mean 6.20 bl/s and standard deviation 1.58 bl/s. (a) What is the probability that an ant's speed in light traffic is faster than 5 bl/s? You may find Table B useful. (Enter your answer rounded to four decimal places.)

Answer:

0.7762

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

Population mean = 6.20 bl/s

Standard deviation = 1.58 bl/s.

x = 5 bl/s

z = 5 - 6.20/1.58

z = -0.75949

The probability that an ant's speed in light traffic is faster than 5 bl/s is P( x > 5)

Probability value from Z-Table:

P(x<5) = 0.22378

P(x>5) = 1 - P(x<5)

= 1 - 22378

= 0.77622

Approximately to 4 decimal places = 0.7762

The probability that an ant's speed in light traffic is faster than 5 bl/s is 0.7762

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If there’s a total of 210 fruit trees, and each tree produces 590 pounds a year, you multiply both those numbers and get 123,900 pounds a year.
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