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MakcuM [25]
3 years ago
6

Help asap please!!!!

Mathematics
1 answer:
Novay_Z [31]3 years ago
5 0
(4,-3) I believe- I got it bc x1 is 3 away from x2 and same with y1 and y2
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What is the vector sum of (7,5) and (13,-5)
kolezko [41]

The vector sum of  two vectors as (7,5) and (13,-5) is given by v_1+v_2=20i+0j which can be written in point form as  (20,0) .

<u>Step-by-step explanation:</u>

Here we have , two vectors as (7,5) and (13,-5) Which are represented as :

v_1=7i+5j\\v_2=13i-5j

Addition of vector is as same as normal addition of integers as :

⇒ v_1+v_2=7i+5j+(13i-5j)

⇒ v_1+v_2=(7+13)i+(5-5)j

⇒ v_1+v_2=20i+0j ,

where i & j are the unit vectors in direction x-axis and y-axis respectively.

The vector sum of  two vectors as (7,5) and (13,-5) is given by v_1+v_2=20i+0j which can be written in point form as  (20,0) .

3 0
4 years ago
Read 2 more answers
In a certain city, 80 percent of the households have cable television and 60 percent of the households have videocassette record
alexandr1967 [171]

Answer:

60,000

Step-by-step explanation:

Let the total people in the household be 100%

If 80 percent of the households have cable television, 60 percent of the households have videocassette recorders and the number of households that have both cable television and videocassette recorders is 'x' percent. The number of households having ONLY cable television will be 80-x while households having ONLY videocassette will be 60-x.

To get x, we will have;

80-x + 60-x + x = 100

140-x = 100

x = 140-100

x = 40

This shows that 40% of the households have both cable television and videocassette.

If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be 40% of 150,000

= 40/100 × 150,000

= 60,000

Number of households that owns both television and videocassette could be from 60,000.

3 0
4 years ago
Read 2 more answers
Which number(s) below belong to the solution set of the equation? Check all that apply . x + 50 = 60
nataly862011 [7]

Answer:

x + 50 = 60

x = 60 - 50

× = 10

this is the answer . It just calculation.

5 0
3 years ago
Solve 110 = 10p. <br><br> p = 12 <br><br> p = 100 <br><br> p = 120 <br><br> p = 11
Temka [501]
You would need to divide both sides by 10.
110=10p
11=p
6 0
4 years ago
Hdc produces microcomputer hard drives at four different production facilities (f1, f2, f3, and f4) hard drive production at f1,
OLEGan [10]
Let D denote the event that an HD is defective, and F_i the event that a particular HD was produced at facility i.

You are asked to compute

\mathbb P(F_2\mid D)
\mathbb P(F_4\mid D)
\mathbb P(D^C)

From the definition of conditional probabilities, the first two will require that you first find \mathbb P(D). Once you have this, part (c) is trivial.

I'll demonstrate the computation for part (a). Part (b) is nearly identical.

(a)
\mathbb P(F_2\mid D)=\dfrac{\mathbb P(F_2\cap D)}{\mathbb P(D)}

Presumably, the facility responsible for producing a given HD is independent of whether the HD is defective or not, so \mathbb P(F_2\cap D)=\mathbb P(F_2)\mathbb P(D)=0.20\times0.015.

Use the law of total probability to determine the value of the denominator:

\mathbb P(D)=\mathbb P(D\mid F_1)+\mathbb P(D\mid F_2)+\mathbb P(D\mid F_3)+\mathbb P(D\mid F_4)

We know each of the component probabilities because they are given explicitly: 0.015, 0.02, 0.01, and 0.03, respectively. So

\mathbb P(D)=0.015+0.02+0.01+0.03=0.075

and thus

\mathbb P(F_2\mid D)=\dfrac{0.2\times0.015}{0.075}=0.04

(b) Similarly,
\mathbb P(F_4\mid D)=\dfrac{0.4\times0.03}{0.075}=0.16

(c)
\mathbb P(D^C)=1-\mathbb P(D)=0.925
6 0
3 years ago
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