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Anna [14]
2 years ago
14

Five whole numbers are written in order.

Mathematics
2 answers:
arsen [322]2 years ago
5 0

Answer:

  • x = 7, y = 8

Step-by-step explanation:

<u>Given numbers in order:</u>

  • 4, 6, x, y, 10

The mean is same as median.

The median is x as middle number. Since x is a whole number, the mean is also a whole number.

x and y can be selected from 7, 8 or 9 (numbers between 6 and 10).

<u>The mean is:</u>

  • (4 + 6 + x + y + 10)/5 =
  • (20 + x + y)/5 =
  • 4 + (x + y)/5

It should be a whole number, so (x + y) should be divisible by 5.

The option is (7 + 8) = 15, and 9 is excluded.

<u>Let's verify:</u>

  • Median = x = 7
  • Mean = 4 + (7 + 8)/5 = 4 + 3 = 7
bija089 [108]2 years ago
3 0
  • 4,6,x,y,10

Mean:-

\\ \sf\longmapsto \dfrac{4+6+x+y+10}{5}

\\ \sf\longmapsto \dfrac{x+y+20}{5}

Median:-

  • n=5
  • M(D)=n+1/2=5+1/2=3rd term

\\ \sf\longmapsto M(D)=x

Compare

\\ \sf\longmapsto \dfrac{x+y+20}{5}=x

\\ \sf\longmapsto x+y+20=5x

\\ \sf\longmapsto 4x=y+20

\\ \sf\longmapsto x=y/4+5

\\ \sf\longmapsto x

Take pair of (7,8)

\\ \sf\longmapsto 7=8/4+5\implies 7=2+5

Verified

Hence

  • x=7
  • y=8

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21. x² +9x - 22<br> Factors of cSum of Factors<br> 7<br> 7<br> ?<br> ?<br> 7<br> 7<br> 9<br> 7
Eduardwww [97]

Answer:

\large \begin{array}{| c | c |}\cline{1-2} \sf Factors\:of\:c & \sf Sum\:of\:Factors\\\cline{1-2} 1, -22 & -21 \\\cline{1-2} -1, 22 & 21\\\cline{1-2} 2, -11 & -9 \\\cline{1-2} -2, 11 & 9 \\\cline{1-2}\end{array}

Step-by-step explanation:

Quadratic equation: ax² + bx + c = 0, where a ≠ 0

x² + 9x - 22

  • a = 1
  • b = 9
  • c = -22

Factors of c (-22):

1 × -22     or     -1 × 22

2 × -11      or     -2 × 11

Sum of factors:

1 + (-22) = -21          or          -1 + 22 = 21

2 + (-11) = -9            or           -2 + 11 = 9

Table:

\large \begin{array}{| c | c |}\cline{1-2} \sf Factors\:of\:c & \sf Sum\:of\:Factors\\\cline{1-2} 1, -22 & -21 \\\cline{1-2} -1, 22 & 21\\\cline{1-2} 2, -11 & -9 \\\cline{1-2} -2, 11 & 9 \\\cline{1-2}\end{array}

Hope this helps!

6 0
2 years ago
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.

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