П

erik [133]

Answer:

6d

Step-by-step explanation

First, you need to multiply 2d by 3, which you'll get 3d.

Next, multiply both numbers (2d and 3d) and you'll have 6d because 2 times 3 equals 6 but, since you have valuables you'd have to put d as the valuables of both numbers then multiply the numbers.

7 0

3 years ago

Jose uses the pattern in rows 1 and 8 in the multiplication table below to find ratios that are equivalent to 1.8. 2 3 6 7 8 9 1

luda_lava [24]

Answer:

it is 20

Step-by-step explanation:

5 0

3 years ago

What is 9b minus 3 over 9 equals 5b plus 5 over 3

Flura [38]

You reduce the fraction which would be 9b-1/3=5b+ 5/3
Then you move the variable to the left side and change its sign 9b-5b-1/3=5/3 then move the constant to the right side 9b-5b=5/3+1/3
Then collect the terms 4b= 5/3 + 1/3 add the fractions 4b = 2
Divide both sides of the equation which is b= 1/2

8 0

3 years ago

Luis walked one-eighth of the distance to school. What part of distance to school does he still need to walk?

klio [65]

Total Distance = x
Luis walked one-eighth of x = x(1/8) = x/8
He still need to walk = Total Distance - Distance covered
He still need to walk = x-x/8
He still need to walk = (8x-x)/8

He still need to walk = 7x/8
He still need to walk = x(7/8)
He still need to walk seven-eighth of the total distance.

6 0

3 years ago

Read 2 more answers

AllElectronics carries 1000 products, P1, . . . , P1000. Consider customers Ada, Bob, and Cathy such that Ada and Bob purchase t

____ [38]

Answer:

The probability that dist(Ada,Bob)>(Ada,Cathy) is very small as there is very large number of range to choose the product ==4.7*10^-9.

Step-by-step explanation:

Given:

Ada ,bob and cathy purchase electronics carries

Ada and bob commonly take 3 products and 7 independently.

And Cathy take 10 products on its own .

To Find:

probability that dist(Ada,bob)>dis(Ada,Cathy)?

Solution:

Using Euclidean distance is distance formula used in coordinate geometry simply known as Distance formula,

this problem is related to Euclidean Distance and Jaccard Similarity in Data mining.

1st calculate probability for x such that ,

$3\leq x\leq 10$ as there are 3 common products.

P( $3\leq x\leq 10$ )

= $\frac{7C(x-3)*990C(10-x)}{997C7}$ .............. where x=3,4,5....10. ..........(equation 1).

Now calculate for each term,we get

When

x=3,P(x=3)=0.95

x=4,P(x=4)= $6.8*10^{-3}$

x=5,P(x=5)= $4.1*10^{-5}$

x=6,P(x=6)=2.1*10^-7

x=7,P(x=7)=8.5*10^-10

x=8,P(x=8)=2.6*10^-12

x=9,P(x=9)=5.2*10^-15

x=10,P(x=10)=5.3*10^-18.

Now calculating the Euclidean distance,

It is distance between two points ,

So there are total of 2 points as Ada and bob

they have 3 products in common

and 7 independent products ,7 Ada and 7 bob

Total of 17 products .

1,2,3,4,5,6..........,16,17.

Consider each product number as distance between them ,

(Suppose 5 product and 1 product distance will be 4)

Similarly,

Suppose Ada is at 3rd number at the 3 product (as they have 3 product same.)

and bob at product 17.

Hence when 3 products are similar distance between Ada and bob will be of 14.

Euclidean distance = $\sqrt{14}$ .

Hence the Jaccard similarity =(Ada intersection Bob)/(Ada union bob)

=3/14

When 4 products are same means both will selected 6 and 6 independent product so that the each one will get 10 products i.e. starting condition should remain same .

Hence now bob will be at 16th term as it will take one more same product in between them

So no of same products will be 4,

Hence Ada will be at 4th term and bob will be at 16

So Euclidean distance = $\sqrt{12}$ .

Similar For Next terms we can conclude as follows:

When

X=5 , dist(ada,bob)= $\sqrt{10}$ ,

X=6,dist(Ada,Bob)= $\sqrt{8}$

X=7,dist(Ada,Bob)= $\sqrt{6}$

X=8,dist(Ada,Bob)= $\sqrt{4}$

X=9,dist(Ada,Bob)= $\sqrt{2}$

X=10,dist(Ada,Bob)= $\sqrt{0}$ .

Now for( Ada and cathy)

Here X ranges different but use same concept as above

Each term analog to the distance between them

Suppose 1st and 3rd term distance will be 2

First calculate

P( $1\leq x\leq 10$ ) as Cathy selects 10 products with no common between them.

P( $1\leq x\leq 10$ )

= $\frac{10Cx*990C(10-x)}{1000C10}$ ..................equation (2)

Calculate for each term As x=1,2,3...8,9,10.

Hence

P(X=1)=9.23*10^-3 P(X=5)=3*10^-11 P(X=9)=3.8*10^-21

P(X=2)=8.4*10^-5 P(X=6)=1.5*10^-13 P(X=10)=3.8*10^-21

P(X=3)=6.9*10^-7 P(X=7)=6.1*10^-16

P(X=4)=4.9*10^-9 P(X=8)=1.9*10^-18

So Ada will have 10 products and Cathy will have 10 products

Namely,

1,2,3,4,5.......18,19,20.

So suppose 1 product is same between them will be ,

both will have 1 product so remaining will be 19 products.

Jaccard similarity =1/19

Distance to reach 1 to 19th product will be 18

So Euclidean distance = $\sqrt{18}$

For next when they will 2 products in same remaining will be 18

Jaccard similarity =2/18

And Distance to reach 2 to 18 th product will be 16

Euclidean distance = $\sqrt{16}$

Similar for other

When

x=3 dist(Ada, Cathy)= $\sqrt{14}$

x=4 dist(Ada, Cathy)= $\sqrt{12}$

x=5 dist(Ada, Cathy)= $\sqrt{10}$

x=6 dist(Ada, Cathy)= $\sqrt{8}$

x=7 dist(Ada, Cathy)= $\sqrt{6}$

x=8 dist(Ada, Cathy)= $\sqrt{4}$

x=9 dist(Ada, Cathy)= $\sqrt{2}$

x=10 dist(Ada, Cathy)= $\sqrt{0}$

This sqrt(0) means both are holding same products hence they are at same point on the graph so distance with itself will be zero.

Now the Probability of distance of dist(Ada,Bob)>dist(Ada,cathy) will be

=multiplying both the probabilities equations (Adding each term probabilities and multiplying )

=Equation(1) *Equation( 2).

=Summation Of P(3≤x≤10)*summation of P(1≤x≤10)

=4.7*10^-9.

In larger number of product event of in large space ,it is difficult( less likely) that they will chose same product .

7 0

3 years ago