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Yuliya22 [10]
3 years ago
7

Slove these 2questions

Mathematics
1 answer:
mart [117]3 years ago
6 0
To find the scale factor divide the bigger measurement by the smaller one. 

You did number 17. 42 / 3 = 14 so your scale factor is 14, option D.

For number 18, 48 / 6 = 8 so option A. 
You might be interested in
Given the vectors u= <2, 1> and v= <5, 4>, determine the components of a vector represented by 2u - 3v
kati45 [8]

Answer:

C.) <-11, -10>

Step-by-step explanation:

Let's define how to work with vectors.

For two vectors:

V = <a, b>

W = <c, d>

The product of a scalar k and a vector is given by:

k*V = k*<a, b> = <k*a, k*b>

And the sum (or difference) of two vectors is given by:

V ± W  = <a, b> ± <c, d> = <a ± c, b ± d>

Now that we know this, we can solve the problem.

Here we have the vectors:

u = <2, 1>

v = <5, 4>

then:

2u - 3*v = 2*<2, 1> - 3*<5, 4>

             = <2*2, 2*1> - <3*5, 3*4>

             = <4, 2> - <15, 12>

             = <4 - 15, 2 - 12>

             = < -11, -10>

Then the correct option is C.

5 0
2 years ago
Plzz help if correct I will mark BRAINLYLIST!!!!
viktelen [127]

Answer:

71

Step-by-step explanation:

82 plus 27=109

180-109=71

7 0
3 years ago
The product of 2 and 5 and then divide 3
Vesnalui [34]
2x5= p divided by 3
2x5=10
10÷3=3.33
so it is 3 1/3
5 0
2 years ago
Read 2 more answers
Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen
k0ka [10]

Let a,b,c be the randomly selected lengths. Without loss of generality, suppose a[tex]P(A + B \ge C) = P(A + B - C \ge 0)

where A,B,C are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)

so that the joint density function is

P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}

where [0,1]^3=[0,1]\times[0,1]\times[0,1] is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

a + b - c = 0

with (a,b,c)\in\Bbb R^3. This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region a+b-c\ge0, since 0+0-1=-1. So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}

5 0
1 year ago
I NEED HELP PLEASE VERY URGENT
I am Lyosha [343]
(X,Y)=(39,69) is the answer
4 0
2 years ago
Read 2 more answers
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