54 ÷ ( 8 - 5 )full explanation too please
2 answers:
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Take the .5 of the 17.5 and convert it to a fractin.
we know .5 is equal to 50% or 1/2
now we have a mixed number
in order to convert it into an improper fraction you need to multiply the whole number by the denominator and then add the numerator all while keeping the denominator constant.
so
we know .5 is equal to 50% or 1/2
now we have a mixed number
in order to convert it into an improper fraction you need to multiply the whole number by the denominator and then add the numerator all while keeping the denominator constant.
so
Answer:
We conclude that the sodium content is same as what the nutrition label states.
Step-by-step explanation:
We are given that the nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium.
The FDA randomly selects 40 packages of Oriental Spice Sauce and determines the sodium content. The sample has an average of 1088.64 milligrams of sodium per package with a sample standard deviation of 234.12 milligrams.
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<u><em>Let </em></u><u><em> = average sodium content.</em></u>
So, Null Hypothesis, :
= 1100 milligrams {means that the sodium content is same as what the nutrition label states}
Alternate Hypothesis, :
1100 milligrams {means that the sodium content is different from what the nutrition label states}
The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about population standard deviation;
T.S. = ~
where, = sample average sodium content = 1088.64 milligrams
s = sample standard deviation = 234.12 milligrams
n = sample of packages of Oriental Spice Sauce = 40
So, <u><em>test statistics</em></u> = ~
= -0.307
The value of z test statistics is -0.307.
<em>Since, in the question we are not given the level of significance so we assume it to be 5%. </em><em>Now, at 0.05 significance level the t table gives critical values of -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test.</em><em> </em>
<em>Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which </em><em><u>we fail to reject our null hypothesis</u></em><em>.</em>
Therefore, we conclude that the sodium content is same as what the nutrition label states.