The answer to your question is,
A. Multiplication
-Mabel <3
The inverse operationof division is multiplication.
Hope this helps you
Brainliest would be appreciated!
- AaronWiseIsBae
Check the picture below.
The student correctly solved the equations given for x. Note that an equation with an unknown variable squared might have two solutions. The way to solve for x alters according to what the equation requires, whether it is adding, subtracting, or dividing.
It seems like the student is trying to solve equations for x. The equations given were all solved correctly. Keep in mind that when an equation contains an unknown variable squared, there could be two solutions, and one or both could be reasonable depending on the problem. For example, consider the equation x² +0.0211x -0.0211 = 0. This could be rearranged to solve for x. Other variables are known unless additional calculations needed if they are not.
Remember that the principle of altering the equation to solve for x is employed, whether we add, subtract or divide by certain values. Like mentioned in the information provided, when dividing by powers of 10, you would move the decimal to the left, corresponding to the number of zeros in the power of ten.
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Answer:
null hypothesis = µ1=µ2=µ3; µ1= popuation mean of inulin µ2 = population mean of fructicoligosaccharide =µ3=population mean of lactulose
alternative hypothesis =µ1≠µ2 ≠µ3
t1= µ-45/s1 / N-1
at the significance level 0.01
t at ᵅ/2 =0.005 and degree of freedom= 35-1=34 is 2.25
t1 = µ-32/s1/N-1
2.25= µ-45/s1/34
= s1/34= µ-45/2.25
=s1 =(µ-45/2.25)*34
t2= µ-32/s2/34
2.25 =µ-32/s2/34
s2/34= µ-32/2.25
s2=µ-32/2.25*34
T= 45-32/s1/sqrt34+s2/sqrt34
t at 0.005 and no of grees of freedom 68 =2.37
2.37=45-32/
or s1/sqrt34+s2/sqrt34 = 13/2.37
s1+s2 = 13/2.37 *5.83
s1+s2= 31.98 or 32
(µ1-45/2.25)*34+µ2-32/2.25*34=32
or µ1+µ2 = 2525
µ1=2525-µ2
µ1 and µ2 are not equal
thus null hypothesis is rejected
conclusion all the three components are not in equal amount in hydrogen production
Explanation : in this experiment we have to prove whether the means of insulin,fructicoligosaccharide and lactulose are equal. so even if we prove that two of them are not equal null hypothesis will be rejected. we use student-t test because we have to compare the means of two population.
(a) For n = 6, CL = 90%,
The degrees of freedom: 5, Critical t-value: 2.571
(b) For n = 21, CL = 98%,
The degrees of freedom: 20, Critical t-value: 2.845
(c) For n = 29, CL = 95%,
The degrees of freedom: 28, Critical t-value: 2.048
(d) For n = 12, CL = 99%,
The degrees of freedom: 11, Critical t-value: 3.106
Use the concept of critical t- value defined as:
A critical value is a number that is used in hypothesis testing to compare to a test statistic and evaluate whether or not the null hypothesis should be rejected. The null hypothesis cannot be rejected if the test statistic's value is less extreme than the crucial value.
(a) Given that,
n = 6 and a confidence level of 90%,
The degrees of freedom are,
n-1 = 6-1
The degrees of freedom = 5.
To find the critical t-value,
Look it up in the t-distribution table using a confidence level of 90% and a degree of freedom of 5.
From the table,
The critical t-value is approximately 2.571.
(b) Given that,
n = 21 and a confidence level of 98%,
The degrees of freedom are,
n-1 = 21-1
The degrees of freedom = 20.
By referring to the t-distribution table with a confidence level of 98% and degrees of freedom of 20,
The critical t-value is approximately 2.845.
(c) Given that,
n = 29 and a confidence level of 95%,
The degrees of freedom are,
n-1 = 29-1
The degrees of freedom = 28
Using the t-distribution table with a confidence level of 95% and degrees of freedom of 28,
The critical t-value is approximately 2.048.
(d) Given that,
n = 12 and a confidence level of 99%,
The degrees of freedom are,
n-1 = 12-1
The degrees of freedom = 11
By consulting the t-distribution table with a confidence level of 99% and degrees of freedom of 11,
The critical t-value is approximately 3.106.
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To find the degrees of freedom and critical t-value for each given sample size and confidence level, we can use the t-distribution and a t-table. The degrees of freedom (df) for each sample is equal to the sample size minus 1. The critical t-value can be found using the t-table with the corresponding degrees of freedom and the confidence level.
To find the degrees of freedom and critical t-value for each given sample size and confidence level, we can use the t-distribution and a t-table. The degrees of freedom (df) for each sample is equal to the sample size minus 1. For example, for (a) n = 6, df = 6 - 1 = 5. The critical t-value can be found using the t-table with the corresponding degrees of freedom and the confidence level.
For (a) n = 6, CL = 90%, the critical t-value is approximately 1.943.
For (b) n = 21, CL = 98%, the critical t-value is approximately 2.861.
For (c) n = 29, CL = 95%, the critical t-value is approximately 2.045.
For (d) n = 12, CL = 99%, the critical t-value is approximately 3.106.
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