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zvonat [6]
2 years ago
13

What is the greatest common factor between 110 and 132​

Mathematics
2 answers:
valentina_108 [34]2 years ago
6 0
The greatest common factor is 22




Explanation:


It is the largest number that goes into both 110 and 132.
Bas_tet [7]2 years ago
4 0

Answer:

the greatest common factor between 110 and 132 is 22

Step-by-step explanation:

hopefully this helps :)

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Find the value of x ​
givi [52]

Answer:

110°

Step-by-step explanation:

In triangle AOB

OA = OB (radii of same circle)

m\angle OAB = m\angle OBA =20\degree

m\angle AOB = 180\degree - (20\degree +20\degree)

m\angle AOB = 180\degree - 40\degree

m\angle AOB = 140\degree

m\widehat {ACB} =m\angle AOB = 140\degree

(measure of minor arc is equal to corresponding central angle)

x\degree = \frac{1}{2} (360\degree - 140\degree)

(inscribed angle theorem)

x\degree = \frac{1}{2} \times 220\degree

\huge\purple {x = 110}

3 0
2 years ago
Which equation could be used to solve for the value of n if n + 2 = 6?
NNADVOKAT [17]

You can use 6-2=4

You just Subtract the 2 from 6 and that gives 4

5 0
3 years ago
What is 32 divided by 526
Marianna [84]

Answer:

0.0608365019

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7 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
Divide (–15x5 – 16x4 + 84x3 – 17x2 – 9x – 15) by (x – 8/5) using synthetic division.
xz_007 [3.2K]
Synthetic division work with the coefficient of the given polynomial expression

We have 
-15    -16   84   -17   -9   -15

and the divisor is:
x - ⁸/₅ = 0
x = ⁸/₅

Refer to the diagram below for the steps of synthetic division

Start by multiplying the first coefficient by the divisor, write the answer under the second coefficient, and then add the two values.

Repeat the steps until we use up all the remaining coefficients

The final values are the coefficients of the quotient and the last value is the reminder


6 0
3 years ago
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