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3 years ago

Count the number of 5-digit numbers (in the decimal system; integers from 10000 to 99999) which are not divisible by 11.

Mathematics

1 answer:

Alexxandr [17]3 years ago

4 0

Answer:

So total number which is not divisible by 11 = 89999-8181 = 81819

Step-by-step explanation:

We have to find the total numbers which are not divisible by 11 from 10000 to 99999

Total number from 10000 to 99999 = 99999-10000+1=90000

Just after 10000 the number which is divisible by 11 is 10010

And just below 99999 the number which is divisible by 11 is 99990

Using AP formula

$T_N=a+(n-1)d$ , here a is first term, d is common difference and n is number of terms

So $99990=10010+(n-1)11$

n-1 = 8180

n= 8181

So total number which is not divisible by 11 = 90000-8181 = 81819

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The diameter of the inscribed circle in a regular hexagon is 4√3 inches long. What is the perimeter of this regular hexagon?

barxatty [35]

Answer:

12√3 inches or 20.785 inches.

Step-by-step explanation:

A regular hexagon can be defined as a polygon with 6 sides.

The formula for the perimeter of a regular hexagon =

6 × the length of the sides of the hexagon.

From the above question, we are told that there is an inscribed circle I'm the hexagon with a diameter of 4√3 inches long

Step 1

Find the radius of the circle

Radius of the circle = 4√3/2 = 2√3 inches

Step 2

The radius of the inscribed circle = Length of one of the sides of a regular hexagon.

Hence, the perimeter of the regular hexagon = 6 × 2√3

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3 years ago

Why do we use congruence?

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For people to comprehend the structure of their world, congruence is a crucial mathematical concept. Young children's daily interactions with congruence enable them to build innate perceptions of this geometric relationship.

Congruent:

If the size and shape of two triangles are same, they are said to be congruent. Three equal sides and identical angles are shared by two congruent triangles with regard to one another. Corresponding Parts of Congruent Triangles are Congruent is referred to by the theorem CPCTC. According to the CPCTC theorem, whenever two triangles are congruent, then each of their corresponding parts are also congruent. In other words, if two or more triangles are congruent, then their corresponding sides and angles are also congruent or equivalent in size.

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3 years ago

A survey said that 3 out of 5 students enrolled in higher education took at least one online course last fall. Explain your calc

marysya [2.9K]

Answer:

a) 60% probability that student took at least one online course

b) 40% probability that student did not take an online course

c) 12.96% probability that all 4 students selected took online courses.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they took at least one online course last fall, or they did not. The probability of a student taking an online course is independent of other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

3 out of 5 students enrolled in higher education took at least one online course last fall.

This means that $p = \frac{3}{5} = 0.6$

a) If you were to pick at random 1 student enrolled in higher education, what is the probability that student took at least one online course?

This is P(X = 1) when n = 1. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{1,1}.(0.6)^{1}.(0.4)^{0} = 0.6$

60% probability that student took at least one online course.

b) If you were to pick at random 1 student enrolled in higher education, what is the probability that student did not take an online course?

This is P(X = 0) when n = 1.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{1,1}.(0.6)^{0}.(0.4)^{1} = 0.4$

40% probability that student did not take an online course

c) Now, consider the scenario that you are going to select random select 4 students enrolled in higher education. Find the probability that all 4 students selected took online courses

This is P(X = 4) when n = 4.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{4,4}.(0.6)^{4}.(0.4)^{0} = 0.1296$

12.96% probability that all 4 students selected took online courses.

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azamat

Step-by-step explanation:

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