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zimovet [89]
3 years ago
9

Given that Ray B A bisects ∠DBC, which statement must be true? m∠ABD = m∠ABC AB ≅ BC B is the midpoint of DC. m∠DBC = 90°

Mathematics
2 answers:
Sloan [31]3 years ago
8 0

Answer:

A

Step-by-step explanation:

Because I said this was the answer. I know all.

nadya68 [22]3 years ago
4 0

Answer:

A.m\angle ABD=m\angle ABC

Step-by-step explanation:

We are given that a  ray BA bisects angle DBC.

We have to find true statement .

Angle bisector property:When a ray bisect any angle then the angles  made by bisection of angle are equal.

When ray BA bisects angle DBC

Then, m\angle ABD=m\angle ABC

By angle bisector property.

Therefore, option A is true.

Answer:A.m\angle ABD=m\angle ABC

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The answer is E. 4x=81
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3 years ago
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What is the domain of the given relation?
Harman [31]

Step-by-step explanation:

Domain is the range of x-values, which is {0, 1, 2}. (D)

6 0
3 years ago
To which set of numbers does the number-5 belong? select all that apply
Lelu [443]
A, C, D are the correct answers.

-5 is neither a whole number nor a natural number. This is because negative numbers are not included in these. Hope this helps! Let me know:)
4 0
3 years ago
In how many unique ways can the seven letters in the word MINIMUM be arranged, if all the letters are used each time
slega [8]

The number of unique ways or permuations to arrange the seven letters in MINIMUM is all the letters are used each time is 420.

According to the given question.

We have a word MINIMUM.

Here, there are 7 letters in "MINIMUM" .

Now, in Minimum the number of letters which are repeated and which are not.

M = 3 times

I = 2 times

N = 1 time

U = 1 time

As, we all know if there is no repetitions in a word which is made of n letters, then we can arrange it by n! ways.

But if there is repetition, we use formula

\frac{n!}{n_{1}! n_{2}!..n_{k}!  }

where, n = n_{1} +n_{2} +n_{3} ...+n_{k}

n_{1} is objects of one type

n_{2} is the objects of two types

n_{k} is the objects of k types

Thereofore, the number of unique ways or permuations to arrange the seven letters in MINIMUM is all the letters are used each time

= 7!/ 3!2!

= 7(6)(5)(4)(3!)/3!(2)(1)

= 7(3)(5)(4)

= 420

Hence, the number of unique ways or permuations to arrange the seven letters in MINIMUM is all the letters are used each time is 420.

Find out more information about number of ways and permuations here:

brainly.com/question/15609044

#SPJ4

4 0
1 year ago
You are planning a survey of starting salaries for recent computer science majors. In the latest survey by the National Associat
Ad libitum [116K]

Answer:

A sample size of at least 228 must be needed.

Step-by-step explanation:

We are given that in the latest survey by the National Association of Colleges and Employers, the average starting salary was reported to be $61,238. Assume that the standard deviation is $3850.

And we have to find that what sample size do we need to have a margin of error equal to $500 with 95% confidence.

As we know that the Margin of error formula is given by;

 <u>Margin of error</u> = Z_\frac{\alpha}{2} \times \frac{\sigma}{\sqrt{n} }

where, \alpha = significance level = 1 - 0.95 = 0.05 and  (\frac{\alpha}{2}) = 0.025.

            \sigma = standard deviation = $3,850

            n = sample size

<em>Also, at 0.025 significance level the z table gives critical value of 1.96.</em>

So, margin of error is ;

                        500=1.96 \times \frac{3,850}{\sqrt{n} }

                         \sqrt{n} = \frac{1.96 \times 3,850}{500}

                          \sqrt{n} = 15.092

 Squaring both sides we get,

                            n = 15.092^{2} = 227.8 ≈ 228

So, we must need at least a sample size of 228 to have a margin of error equal to $500 with 95% confidence.

4 0
3 years ago
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