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blsea [12.9K]
3 years ago
5

Plzzzz help me with one and two I don’t understand

Mathematics
1 answer:
Ratling [72]3 years ago
7 0
What this question simply asks about is you write the numbers it gave you in only one digit by 10 to the power of n. Where n represents the number of shifts of the decimal point.
And to get this number to be one digit, then you have to approximate.
To approximate; you should always consider that less than 5 counts zero and 5 or more counts+1.
Example:
5.3 is approximately 5
5.5 is approx. 6
5.9 is also approx. 6
and so on.
So the final answers are;
1) 4*10^13
2) 5*10^-11
Please note that 10^positive integer will cause the decimal point to move to the right which means a number greater than 1, usually, while 10^negative integer means number smaller than 1, usually.
Hope this helps.
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A student carried out an experiment to determine the amount of vitamin C in a tablet sample. He performed 5 trials to produce th
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There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

P-value = 0.166.

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

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Then, we can perform the hypothesis t-test for the mean.

The claim is that the amount of vitamin C in a tablet sample is different from 500 mg.

Then, the null and alternative hypothesis are:

H_0: \mu=500\\\\H_a:\mu< 500

The significance level is 0.05.

The sample has a size n=5.

The sample mean is M=496.8.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.5.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6.5}{\sqrt{5}}=2.907

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{496.8-500}{2.907}=\dfrac{-3.2}{2.907}=-1.1

The degrees of freedom for this sample size are:

df=n-1=5-1=4

This test is a left-tailed test, with 4 degrees of freedom and t=-1.1, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.166) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

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