I need this ASAP A= Y=6+xB =y=6-xC= y=x-6D= y=6x
1 answer:
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Answer:
The claim on the M&M’s website is not true.
Step-by-step explanation:
A Chi-square test for goodness of fit will be used in this case.
The hypothesis can be defined as:
<em>H</em>₀: The observed frequencies are same as the expected frequencies.
<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.
The test statistic is given as follows:
Here,
= Observed frequencies
= Expected frequency.
The chi-square test statistic value is, 14.433.
The degrees of freedom is:
df = <em>k</em> - 1 = 6 - 1 = 5
Compute the <em>p</em>-value as follows:
*Use a Chi-square table.
The significance level is, <em>α</em> = 0.05.
p-value = 0.013 < <em>α</em> = 0.05.
So, the null hypothesis will be rejected at 5% significance level.
Thus, concluding that the claim on the M&M’s website is not true.
<em>Solve: </em>
Start by isolating the x-variable as much as you can:
Now, if we take the log with base 13 of both sides, then the left hand side will cancel out really easily. This is based on the concept of inverse operations. For example, when we take the square of a square root, we get our initial value. This is because the two operations are inverses of each other, and basically undoes each other.
Logarithms and exponentials are the inverse operations of each other. If we want to extrapolate the power of an exponential, we take the logarithmic function of it, and vice versa.
In this case, we are saying what power of '13' will produce us with 13ˣ, and the only answer will be x itself.
Start by isolating the x-variable as much as you can:
Now, if we take the log with base 13 of both sides, then the left hand side will cancel out really easily. This is based on the concept of inverse operations. For example, when we take the square of a square root, we get our initial value. This is because the two operations are inverses of each other, and basically undoes each other.
Logarithms and exponentials are the inverse operations of each other. If we want to extrapolate the power of an exponential, we take the logarithmic function of it, and vice versa.
In this case, we are saying what power of '13' will produce us with 13ˣ, and the only answer will be x itself.