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3 years ago

6.A = 23 × 3 × 5 B = 22 × 3 × 52

Mathematics

1 answer:

frez [133]3 years ago

6 0

Given:

The two numbers are

$A=23\times 3\times 5$

$B=22\times 3\times 52$

To find:

The highest common factor (HCF) of A and B

Solution:

We have,

$A=23\times 3\times 5$ ...(i)

$B=22\times 3\times 52$

All the factors of A are prime but the factors of B are not prime. So, it can be written as

$B=2\times 11\times 3\times 13\times 2\times 2$ ...(ii)

From (i) and (ii), it is clear that 3 is the only common factor of A and B. So,

$HCF(A,B) = 3$

Therefore, the highest common factor (HCF) of A and B is 3.

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One of ten different prizes was randomly put into each box of a cereal. If a family decided to buy this cereal until it obtained

Reptile [31]

Answer:

29.29

Step-by-step explanation:

The first trial among a distribution of independent trials with a constant success probability follows a geometric probability.

The expected value of a negative binomial is the reciprocal of its success probability p.

E(X)=1/p

At the first draw, we only selected one of the 10 prizes and so 1 draw gives a success

E(X)=1/p

E(X1)=1

After the first prize is drawn, we are interested in the first time (success), that we draw another prize that is different (p=9/10)

E(X2)=1/9÷10

E(X2)= 10/9

After the first prizes have being drawn, we now interested in the first time success that we draw a different prize from the first 2

E(X3)=1/8÷10

E(X3)=10/8

E(X3)=5/4

After the first three prizes have been drawn, we are now interested in the success of a first time draw for the 4th different prize

E(X4)=1/7÷10

E(X4)=10/7

Fifth different prize

E(X5)=1/6÷10

E(X5)=10/6

E(X5)=5/3

Sixth different prize

E(X6)=1/5÷10

E(X6)=10/5

E(X6)=2

Seventh different prize

E(X7)=1/4÷10

E(X7)=10/4

E(X7)=5/2

Eighth different prize

E(X8)=1/3÷10

E(X8)=10/3

Ninth different prize

E(X9)=1/2÷10

E(X9)=10/2

E(X9)=5

After the 9 different prizes have been drawn, we are interested in the first time we will draw the tenth different prize

E(X10)=1/1÷10

E(X10)=10

Add up all corresponding values;

E(X)=E(X1)+E(X2)+E(X3)+E(X4)+E(X5)+E(X6)+E(X7)+E(X8)+E(X9)+E(X10)

E(X)=1+10/9+5/4+10/7+5/3+2+5/2+10/3+5+10

E(X)= 29.29

Note: Those values in E(X), should be written the way it will in standard algebra ie they should be small.

4 0

3 years ago

Joe’s annual income has been increasing each year by the same dollar amount the first year his income was 19,500. 8 years later

Anna007 [38]

Answer:

The question is either poorly constructed or it is a trick question. We do not know "In what year was his income 30,300" because the years correlating to his income are not disclosed. Even assuming the question was meant to be posed as "how many years later was his income 30,300" it is still flawed.

Step-by-step explanation:

To arrive at Joe's annual income increase, first subtract 19,500 from 27,900 to get 8400, which is his overall increase over the 8 years. Now divide 8400 by 8. You will get 1050. Beginning with his income of 27,900, add 1,050 to each successive income amount and you will get the following progression:

27,900, 28,950, 30,000, 31,050, 32,100, 33,150, 34,200, 35,250, 36,300.

Therefore, two years after his income was 27,900 it would be 30,000 and 6 years after that it would be 36,300. It would never be 30,300.

6 0

3 years ago

What is down payment with regard to buying a house?

Lyrx [107]

The down payment is an initial payment made when something is bought on credit.

It usually depends on the type of house or any other other form of object.

7 0

3 years ago

Without calculating, which has a bigger volume. A cube that has a length, width, and height of 18 m. Or a sphere with a radius o

Evgesh-ka [11]

Weird. A period appears above this... huh.

Answer:

[Th]e cube has a greater value.

Step-by-step explanation:

What the word problem really wants us to get [is ]the question of 'Which is greater, A=6a^2 when 'a' [is] 18 or A=4 $\pi$ r^2 when r = 9? And here's how to solve that.

Starting with the[ c]ube we have A=6a^2. A bit t[o]o simple, right?

A=6(18)^2 Substitute numbers.

A=6(324) Solve ex[p]onents.

A=1944 Mult[i]ply.

So w[e] know that the cube is 1944 meters cube[d ] in area. But what about the more [f]ormidable sphere? Fo[r] it we need a slightly m[o]re co[m]plicated formula, A=4 $\pi$ r^2. However, instead of using the real pi I will be rounding to 3.14, since we have no calculator so anything more would take way too long and fry your[ bra]in.

A=4(3.14)(9)^2 Subst[i]tute numbers.

A=4(3.14)(81) Solve expone[n]ts.

A=12.56(81) Multip[ly].

A=1017.36 Multiply again[.]

Now, since I'm sure all of us can count, we know that 1944 is greater than 1017.36. Or in other words, the cube is bigger than the sphere.

And PLEASE don't copy this guys. Make your own iteration. Change it up!

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3 years ago

What is the answer to this

Gwar [14]

The slope of this line is 0, it would be undefined if the line were vertical

4 0

3 years ago

Read 2 more answers