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4 years ago

50 POINTS!!!!! Name an angle that is complementary to ∠BOC.

Mathematics

2 answers:

Stolb23 [73]4 years ago

7 0

The complementary angle to ∠BOC is ∠COD.

BigorU [14]4 years ago

6 0

Angle COD a complementary angle is an angle that combines with another to create a 90 degree angle

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0.198 is what percent of 6?<br> 7

oksano4ka [1.4K]

Answer:

3.3%

Step-by-step explanation:

.198/6= .033

.033x100= 3.3%

7 0

3 years ago

Read 2 more answers

Can someone help with this?? :)))

Y_Kistochka [10]

Answer:

5.3

You split them up into smaller root numbers and add them together

6 0

2 years ago

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Consider the discussion in our Devore reading in this unit involving an important distinction between mean and median that uses

Levart [38]

Answer:

Step-by-step explanation:

A trimmed mean is a method of averaging that removes a small designated percentage of the largest and smallest values before calculating the mean. After removing the specified observations, the trimmed mean is found using a standard arithmetic averaging formula. The use of a trimmed mean helps eliminate the influence of data points on the tails that may unfairly affect the traditional mean.

trimmed means provide a better estimation of the location of the bulk of the observations than the mean when sampling from asymmetric distributions;

the standard error of the trimmed mean is less affected by outliers and asymmetry than the mean, so that tests using trimmed means can have more power than tests using the mean.

if we use a trimmed mean in an inferential test , we make inferences about the population trimmed mean, not the population mean. The same is true for the median or any other measure of central tendency.

I can imagine saying the skewness is such-and-such, but that's mostly a side-effect of a few outliers, the fact that the 5% trimmed skewness is such-and-such.

I don't think that trimmed skewness or kurtosis is very much used in practice, partly because

If the skewness and kurtosis are highly dependent on outliers, they are not necessarily useful measures, and trimming arbitrarily solves that problem by ignoring it.

Problems with inconvenient distribution shapes are often best solved by working on a transformed scale.

There can be better ways of measuring or more generally assessing skewness and kurtosis, such as the method above or L-moments. As a skewness measure (mean ? median) / SD is easy to think about yet often neglected; it can be very useful, not least because it is bounded within [?1,1][?1,1].

i expect to see the optimum point in that process at some value between the mean and median.

3 0

3 years ago

Simplifying rational expressions <br> x^2-16/x^2+6x+8 = x-4/x+2<br> x^2-x-6/x^2-3x-10 = x-3/x-5

enot [183]

$\dfrac{x^2-16}{x^2+6x+8}$

Decompose the numerator and denominator into multipliers

To simplify the numerator we use the formula of difference of squares

$x^2-y^2=(x-y)(x+y)$

$x^2-16=(x-4)(x+4)$

To decompose the denominator into multipliers solve the square equation

$x^2+6x+8=0\\D=6^2-4*8=4=2^2\\x_1=\dfrac{-6+2}{2} =-2\\x_2=\dfrac{-6-2}{2} =-4$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

$(x-(-2))(x-(-4))=(x+2)(x+4)$

After simplifying the numerator and denominator we get a fraction

$\dfrac{(x-4)(x+4)}{(x+2)(x+4)}=\dfrac{x-4}{x+2}$ , so

$\dfrac{x^2-16}{x^2+6x+8}=\dfrac{(x-4)(x+4)}{(x+2)(x+4)}=\dfrac{x-4}{x+2}$

$\dfrac{x^2-x-6}{x^2-3x-10}$

Decompose the numerator and denominator into multipliers

To decompose the numerator into multipliers solve the square equation

$x^2-x-6=0\\D=(-1)^2-4*(-6)=25=5^2\\x_1=\dfrac{1+5}{2} =3\\x_2=\dfrac{1-5}{2} =-2$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

$(x-3)(x-(-2))=(x-3)(x+2)$

To decompose the denominator into multipliers solve the square equation

$x^2-3x-10=0\\D=(-3)^2-4*(-10)=49=7^2\\x_1=\dfrac{3+7}{2} =5\\x_2=\dfrac{3-7}{2} =-2$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

$(x-5)(x-(-(-2))=(x-5)(x+2)$

After simplifying the numerator and denominator we get a fraction

$\dfrac{(x-3)(x+2)}{(x-5)(x+2)}=\dfrac{x-3}{x-5}$ , so

$\dfrac{x^2-x-6}{x^2-3x-10}=\dfrac{(x-3)(x+2)}{(x-5)(x+2)}=\dfrac{x-3}{x-5}$

Hello from Russia:^)

6 0

3 years ago

Which statement is not used to prove that ΔLKM is similar to ΔNOM?

mel-nik [20]

Answer:

Angle K is congruent to itself, due to the reflexive property

Step-by-step explanation:

Given: ΔLKM is similar to ΔNOM

To find: statement that is not used to prove that ΔLKM is similar to ΔNOM

Solution:

Two figures are said to be congruent if they have same shape and same size.

Two figures are said to be similar if they have same shape but different size.

If two figures are similar then their corresponding angles are equal.

As ΔLKM $\sim$ ΔNOM,

$\angle L=\angle N\,,\,\angle K=\angle O\,,\,\angle M=\angle M$

So, statement that can be used to prove that ΔLKM is similar to ΔNOM is ''Angle K is congruent to itself, due to the reflexive property''

4 0

3 years ago