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DENIUS [597]
3 years ago
8

Which of the following points is an equal distance (equidistant) from a(2, −2) and b(−4, 6)?

Mathematics
1 answer:
lesantik [10]3 years ago
6 0
The point equidistant from a(2, -2) and b(-4, 6)  is the midpoint

For (x₁, y₁) and (x₂, y₂) the midpoint is given by:

 ( (x₁ + x₂)/2 ,  (y₁ + y₂)/2 )

Midpoint = ( (2 + -4)/2 , (-2 + 6)/2 )

               =  ( (2 - 4)/2 ,  (6 - 2)/2) ) = ( -2/2,  4/2) = (-1,  2)

So the point equidistant from the two points is (-1, 2).
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Find the reciprocals of the following numbers: 8, 2/7 , 1 1/5 , 0.3.
Phoenix [80]
1/8, 7/2, 5/6, 10/3

just swap numerator and denominator for each
8 0
3 years ago
14x+10y=72 14x+4y=96 Solve the following system of equations using any method
Anarel [89]
    14x + 10y = 72
    14x +   4y = 96
                6y = -24
                 6       6
                  y = -4
    14x + 10y = 72
14x + 10(-4) = 72
       14x - 40 = 72
             + 40 + 40
              14x = 112
               14      14
                  x = 8
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5 0
3 years ago
What should I buy? A study conducted by a research group in a recent year reported that of cell phone owners used their phones i
Llana [10]

Answer:

The probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7886.

<em />

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of 14 cell phone owners is studied. Round the answers to at least four decimal places. What is the probability that seven or more of them used their phones for guidance on purchasing decisions? </em>

We can model this as a binomial random variable, with p=0.57 and n=14.

P(x=k)=\dbinom{n}{k} p^{k}q^{n-k}

a) We have to calculate the probability that seven or more of them used their phones for guidance on purchasing decisions:

P(x\geq7)=\sum_{k=7}^{14}P(x=k)\\\\\\

P(x=7)=\dbinom{14}{7} p^{7}q^{7}=3432*0.0195*0.0027=0.1824\\\\\\P(x=8) = \dbinom{14}{8} p^{8}q^{6}=3003*0.0111*0.0063=0.2115\\\\\\P(x=9) = \dbinom{14}{9} p^{9}q^{5}=2002*0.0064*0.0147=0.1869\\\\\\P(x=10) = \dbinom{14}{10} p^{10}q^{4}=1001*0.0036*0.0342=0.1239\\\\\\P(x=11) = \dbinom{14}{11} p^{11}q^{3}=364*0.0021*0.0795=0.0597\\\\\\P(x=12) = \dbinom{14}{12} p^{12}q^{2}=91*0.0012*0.1849=0.0198\\\\\\P(x=13) = \dbinom{14}{13} p^{13}q^{1}=14*0.0007*0.43=0.004\\\\\\

P(x=14) = \dbinom{14}{14} p^{14}q^{0}=1*0.0004*1=0.0004\\\\\\

P(x\geq7)=0.1824+0.2115+0.1869+0.1239+0.0597+0.0198+0.004+0.0004\\\\P(x\geq7)=0.7886

7 0
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Translate the sentence into an equation.
rodikova [14]
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Nady [450]
I believe 25 has to be added as
(x*5)^2 = x^2 + 10x + 25
5 0
3 years ago
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