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lidiya [134]
3 years ago
13

The atomic weight of an unknown element,in atomic mass units (amu),is approximately 20% less than that of calcium. The atomic we

ight of calcium is 40 amu. What is the approximate weight,in amu,of the unknown element?
Mathematics
2 answers:
romanna [79]3 years ago
5 0
I don't know what ur talking about I'm in 6th grade LOL

Mnenie [13.5K]3 years ago
4 0
Amu of unknown element = 40 - (20% * 40)
= 40 - 8
= 38 amu

therefore, the approximate weight of the unknown element is 38 amu.

hope this helps.
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Answer:

The null and alternative hypothesis for this study are:

H_0: \mu=18500\\\\H_a:\mu< 18500

The null hypothesis is rejected (P-value=0.004).

There is enough evidence to support the claim that the average cost of tuition plus room and board at small private liberal arts colleges is  less than $18,500 per term.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the average cost of tuition plus room and board at small private liberal arts colleges is  less than $18,500 per term.

Then, the null and alternative hypothesis are:

H_0: \mu=18500\\\\H_a:\mu< 18500

The significance level is 0.05.

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The sample mean is M=18200.

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We can calculate the standard error as:

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This test is a left-tailed test, so the P-value for this test is calculated as:

P-value=P(z

As the P-value (0.004) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the average cost of tuition plus room and board at small private liberal arts colleges is  less than $18,500 per term.

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