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mel-nik [20]
3 years ago
14

With proof please, i need to know how many dozen sugar cookies they should prepare

Mathematics
1 answer:
ipn [44]3 years ago
5 0
40 dozens because 40 dozens would make 480 and 30 dozens will only make 360 and you need 400.
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Notice that the alternative hypothesis is a two-tailed test. Suppose ttest_ind method from scipy module is used to perform the t
Korvikt [17]

Answer:

The <em>p</em>-value of the test is 0.0512.

Step-by-step explanation:

The <em>p</em>-value of a test is well-defined as per the probability, [under the null hypothesis (H₀)], of attaining a result equivalent to or more extreme than what was the truly observed value of the test statistic.

In this case the output of the t-test_ind method from scipy module is provided as follows:

Output = (-1.99, 0.0512)

The first value within the parentheses is the test statistic value.

So the test statistic value is, -1.99.

And the second value within the parentheses is the <em>p</em>-value of the test.

So the <em>p</em>-value of the test is 0.0512.

6 0
3 years ago
25 + 4^2 please help<br>​
maksim [4K]
4^2 equals 16. 16+25=41
The answer would be 41
8 0
3 years ago
Read 2 more answers
Dr. Pagels is a mammalogist who studies meadow and common voles. He frequently traps the moles and has noticed what appears to b
Alina [70]

Answer:

Null hypothesis = H₀ = There food preferences among vole species are independent of one another.

Alternate hypothesis = H₁ = There is a relationship between voles and food preference.

Expected meadow vole/apple slices = 29.983051

Expected common vole/apple slices = 28.016949

Expected meadow vole/peanut butter-oatmeal = 31.016949

Expected common vole/peanut butter-oatmeal = 28.983051

Chi-square value = χ² = 2.154239

Degree of freedom = 1

Critical value = 3.841

χ² < Critical value

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a relationship between voles and food preference.

Step-by-step explanation:

He frequently traps the moles and has noticed what appears to be a preference for a peanut butter-oatmeal mixture by the meadow voles vs apple slices are usually used in traps, where the common voles seem to prefer the apple slices.

So he conducted a study where he used a peanut butter-oatmeal mixture in half the traps and the normal apple slices in his remaining traps to see if there was a food preference between the two different voles.

Null hypothesis = H₀ = There food preferences among vole species are independent of one another.

Alternate hypothesis = H₁ = There is a relationship between voles and food preference.

Data collected by Dr. Pagels:

                                              meadow voles     common voles      Row Total

apple slices                                     26                          32                      58

peanut butter-oatmeal                   35                          25                     60

Column Total                                   61                          57                     118

Where 118 is the grand total.

The expected number is given by

Expected = (row total)×(column total)/grand total

Expected meadow vole/apple slices = 58×61/118

Expected meadow vole/apple slices = 29.983051

Expected common vole/apple slices = 58×57/118

Expected common vole/apple slices = 28.016949

Expected meadow vole/peanut butter-oatmeal = 60×61/118

Expected meadow vole/peanut butter-oatmeal = 31.016949

Expected common vole/peanut butter-oatmeal = 60×57/118

Expected common vole/peanut butter-oatmeal = 28.983051

The chi-square statistic value is given by

χ² = Σ(Observed - Expected)²/Expected

χ² = (26 - 29.983051)²/29.983051 + (32 - 28.016949)²/28.016949 + (35 - 31.016949)²/31.016949 + (25 - 28.983051)²/28.983051

χ² = 2.154239

The degrees of freedom is given by

DoF = (row - 1)×(col - 1)

For the given case, we have 2 rows and 2 columns

DoF = (2 - 1)×(2 - 1)

DoF = 1

The given level of significance = 0.05

The critical value from the chi-square table at α = 0.05 and DoF = 1 is found to be

Critical value = 3.841

Conclusion:

Reject H₀ If χ² > Critical value

We reject the Null hypothesis If the calculated chi-square value is more than the critical value.

For the given case,

χ² < Critical value

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a relationship between voles and food preference.

8 0
3 years ago
A+b-c=180 Solve the equation
VladimirAG [237]
You can do it in multiple ways but mine is....     140+120-160=180
3 0
3 years ago
Read 2 more answers
What does it equal? Because I'm really confused
ruslelena [56]
It equals to 3/13
this is easy to solve because the denominator is already the same. all you'll need to do is add the the numerators:
2+1=3
and your answer will be 3/13
5 0
3 years ago
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