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3 years ago

-8 equals x + 11 / -2

Mathematics

1 answer:

slava [35]3 years ago

5 0

If the equation is $-8=x+\frac{11}{-2}$ :

$-8=x+\frac{11}{-2}\\-8=x-\frac{11}{2}\\-8+\frac{11}{2}=x-\frac{11}{2}+\frac{11}{2}\\-\frac{16}{2}+\frac{11}{2}=x\\x=-\frac{5}{2}=-2\frac{1}{2}$

If the equation is $-8=\frac{x+11}{-2}$ :

$-8=\frac{x+11}{-2}\\-8(-2)=\frac{x+11}{-2}(-2)\\16=\frac{-2(x+11)}{-2}\\16=x+11\\16-11=x+11-11\\x=5$

Hopefully the math was self-explanatory. If it is not, please feel free to ask for a step-by-step, in depth explanation, and I will edit my answer to accommodate.

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ivolga24 [154]

Answer:

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:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{5}(490+502+505+495+492)\\\\\\M=\dfrac{2484}{5}\\\\\\M=496.8\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}((490-496.8)^2+(502-496.8)^2+(505-496.8)^2+(495-496.8)^2+(492-496.8)^2)}\\\\\\s=\sqrt{\dfrac{166.8}{4}}\\\\\\s=\sqrt{41.7}=6.5\\\\\\$

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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Answer:

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Answer:

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