15 points. Need help with this math test

A shop keeper has offered an item for sale. Its label price is Rs . It is not sold in one-month period and after one month its l

BigorU [14]

Answer:

x = Rs 20,833.33

the value of x is Rs 20,833.33

Step-by-step explanation:

Let x,y and z represent the price of the item initially, after one month and after two months respectively.

Given that;

after one month its label price is reduced by 20%

y = x - 20% of x

y = x - 0.20x

y = 0.80x ........1

after 2 months its reduced price is further reduced by 10% and then sold it for Rs 15000.

z = y - 10% of y

z = y - 0.10y

z = 0.90y ........2

Substituting equation 1 into 2;

z = 0.90(0.80x)

z = 0.72x

Also z = Rs 15000

So,

z = 0.72x = Rs 15000

0.72x = Rs 15000

x = Rs 15000/0.72

x = Rs 20833.33333333

x = Rs 20,833.33

the value of x is Rs 20,833.33

5 0

3 years ago

Explain how to estimate the lateral area of a right cone with radius 5 cm and slant height 6 cm. Is your estimate an underestima

Iteru [2.4K]

Answer:

The answer in the procedure

Step-by-step explanation:

we know that

The lateral area of a cone is equal to

$LA=\pi rl$

where

r is the radius of the base

l is the slant height

we have

$r=5\ cm$

$l=6\ cm$

assume $\pi =3.14$

substitute the values

$LA=(3.14)(5)(6)=94.2\ cm^{2}$

This value is an underestimate, because the assumed pi value is less than the real value

assumed value $\pi =3.14$

real value $\pi =3.1415926536...$

8 0

3 years ago

Im bout to do it .......................

ikadub [295]

Answer:

Do what exactly?.....

8 0

2 years ago

Read 2 more answers

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.

3 0

3 years ago

WILL MARK BRAINLIEST

Aneli [31]

Answer:

Table #4, with the r and the s

Step-by-step explanation:

It shows some kind of pattern

4 0

2 years ago