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Yuri [45]
3 years ago
12

A molecule of sodium carbonate contains 2 atoms of sodium to every 3 atoms of oxygen. Could a compound containing 12 atoms of so

dium and 15 atoms of oxygen be sodium carbonate ? Explain
Mathematics
1 answer:
lys-0071 [83]3 years ago
7 0
Ask this question in the science category xoxo
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Identify the transformation performed on the figure below
pickupchik [31]

A) Translation

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3 years ago
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Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is μd =
swat32

Answer:

Test statistic, t_{s} = -0.603 (to 3 dp)

Step-by-step explanation:

Deviation, d = x -y

Sample mean for the deviation

\bar{d} = \frac{\sum x-y}{n}

\bar{d} = \frac{(28-6) + (31-27)+(20-26)+(25-25)+(28-29)+(27-32)+(33-33)+(35-34)}{8} \\\bar{d} = -0.625

Standard deviation: SD = \sqrt{\frac{\sum d^{2} - n \bar{d}^2}{n-1}  }

\sum d^{2}  = (28-26)^2 + (31-27)^2 +(20-26)^2 +(25-25)^2 +(28-29)^2 +(27-32)^2 +(33-33)^2 +(35-34)^2\\\sum d^{2}  = 63

SD = \sqrt{\frac{63 - 8 *  (-0.625)^2}{8-1}  }

SD =2.93

Under the null hypothesis, the formula for the test statistics will be given by:

t_{s} = \frac{ \bar{d}}{s_{d}/\sqrt{n}  } \\t_{s} = \frac{- 0.625}{2.93/\sqrt{8}  }

t_{s} = -0.6033

6 0
3 years ago
When 6 times a number is increased by 11 the result is 16 less than 9 times the number. Find the number.
scZoUnD [109]

Answer:

Brainly.in

Question

When <u>6</u><u> </u><u>times</u><u> </u><u>a</u><u> </u><u>number</u><u> </u> is increased by 11 the result is 16 less than 9 times the number. Find the number.

Answer · 7 votes

Answer:6x+11=9x-163x=27X=9 Please mark it brainliest

Step-by-step explanation:

Please mark it brainliest

5 0
2 years ago
Applied to the graph
lana [24]

Answer:

Using the graph of f(x) and g(x), where g(x) = f(kx), determine the value of k. A: 4 B: 1/4 C:-1/4 D:-4 "

Step-by-step explanation:

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3 years ago
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504,047 written in expanded form
Nutka1998 [239]

Answer:

500,000 + 4,000 + 40 + 7

<h2>BRAINLIEST please if this helped!</h2>
5 0
2 years ago
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