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3 years ago

15

A central purple circle labeled O has 2 concentric rings around it. The inner ring has 2 small green spheres. The outer ring has

6 small green spheres. How many valence electrons does this oxygen atom have?

2 answers:

natulia [17]3 years ago

8 0

Answer:

Its C

Step-by-step explanation:

It is 6 because there are 6 dots on the outer circle and those are the valence electrons. Also I just did it on Edge. Hope this helps.

2 years ago

No, it does not, because the octet rule says that an atom needs to have eight electrons in its valence shell to be stable. The exceptions are hydrogen and helium, which need only two electrons. This atom has only one electron.

Anonymus2 years ago

0 0

No, it does not, because the octet rule says that an atom needs to have eight electrons in its valence shell to be stable. The exceptions are hydrogen and helium, which need only two electrons. This atom has only one electron.

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Draw 6 circles.<br> Make 3 groups.<br> Shade 1 group.

krek1111 [17]

Answer:

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Step-by-step explanation:

2 circles in each group, shade in one group of 2.

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3 years ago

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anzhelika [568]

The function was moved two units to the left (a)

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3 years ago

Standard Error from a Formula and a Bootstrap Distribution Sample A has a count of 30 successes with and Sample B has a count of

tia_tia [17]

Answer:

Using a formula, the standard error is: 0.052

Using bootstrap, the standard error is: 0.050

Comparison:

The calculated standard error using the formula is greater than the standard error using bootstrap

Step-by-step explanation:

Given

Sample A Sample B

$x_A = 30$ $x_B = 50$

$n_A = 100$ $n_B =250$

Solving (a): Standard error using formula

First, calculate the proportion of A

$p_A = \frac{x_A}{n_A}$

$p_A = \frac{30}{100}$

$p_A = 0.30$

The proportion of B

$p_B = \frac{x_B}{n_B}$

$p_B = \frac{50}{250}$

$p_B = 0.20$

The standard error is:

$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * (1 - 0.30)}{100} + \frac{0.20* (1 - 0.20)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * 0.70}{100} + \frac{0.20* 0.80}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.21}{100} + \frac{0.16}{250}}$

$SE_{p_A-p_B} = \sqrt{0.0021+ 0.00064}$

$SE_{p_A-p_B} = \sqrt{0.00274}$

$SE_{p_A-p_B} = 0.052$

Solving (a): Standard error using bootstrapping.

Following the below steps.

Open Statkey
Under Randomization Hypothesis Tests, select Test for Difference in Proportions
Click on Edit data, enter the appropriate data
Click on ok to generate samples
Click on Generate 1000 samples ---- <em>see attachment for the generated data</em>

From the randomization sample, we have:

Sample A Sample B

$x_A = 23$ $x_B = 57$

$n_A = 100$ $n_B =250$

$p_A = 0.230$ $p_A = 0.228$

So, we have:

$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.23 * (1 - 0.23)}{100} + \frac{0.228* (1 - 0.228)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.1771}{100} + \frac{0.176016}{250}}$

$SE_{p_A-p_B} = \sqrt{0.001771 + 0.000704064}$

$SE_{p_A-p_B} = \sqrt{0.002475064}$

$SE_{p_A-p_B} = 0.050$

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malfutka [58]

Answer: 6f (3+5x)

Step-by-step explanation:

18f + 30fx by its gcf

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masha68 [24]

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So yes he had enough to buy all the stuff and still had a little bit leftover!

Hope this helps!

Step-by-step explanation:

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3 years ago