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bagirrra123 [75]
3 years ago
9

A toy manufacturer uses 90 pints of plastic to make 600 action figures. If they use 810 pints of plastic in an 8 hour shift, how

many action figures are made per hour?
Mathematics
1 answer:
lutik1710 [3]3 years ago
8 0

Answer:

The number per hour = 5400 ÷ 8 = 675 action figures

Step-by-step explanation:

* Lets explain how to solve the problem

- A toy manufacturer uses 90 pints of plastic to make 600 action figures

- they use 810 pints of plastic in an 8 hour shift

- To find how many action figures are made per hour we must to find

  how many action figures in an 8 hour shift and then find the number

  of the action figures per hour

∵ The toy manufacture  uses 90 pints of plastic to make 600 action

   figures

∵ They use 810 pints

- Lets find how many action figures are made by this numbers of pints

∴ 90/600 = 810/x ⇒ x is the number of action figures

- Use the cross multiplication to find x

∴ 90 x = 810 × 600

∴ 90 x = 486000 ⇒ divide both sides by 90

∴ x = 5400

∴ They will use 810 pints to make 5400 action figures

∵ They use 810 pints in an 8 hour shift

∴ They make 5400 action figures in 8 hours

- To find the number of the action figures per hour divide the number

  of them in 8 hours by 8

∴ The number per hour = 5400 ÷ 8 = 675 action figures

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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.

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