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dybincka [34]
3 years ago
11

Which digit in the following number is one that determines its precision? 34.867​

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
6 0

6 should be the answer. Because 34 has no meaning because it's behind the dot. 67 has no meaning because the are after the middle. Which means it is the middle to be precision.

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For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?
Bond [772]
Given that r=\theta, then r'=1

The slope of a tangent line in the polar coordinate is given by:

m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}

Thus, we have:

m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}



Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

\sin\theta+\theta\cos\theta=0 \\  \\ \theta\cos\theta=-\sin\theta \\  \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta

Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

</span><span>θ = 0

</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488



Part B:

For vertical tangent lines, \frac{1}{m} =0

Thus, we have:

\cos\theta-\theta\sin\theta=0 \\  \\ \Rightarrow\theta\sin\theta=\cos\theta \\  \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta

</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

</span>θ = <span>4.91718592528713</span>
3 0
3 years ago
Help please uh thanks
Triss [41]
12•1
13•6
12/78 divide both numbers by 6 to simplify
Answer- 2/13
6 0
3 years ago
Read 2 more answers
HaLp PwEaSe <br>what is the least common denominator(LDC) of 5/6 and 11/4 hurry pwease
AleksAgata [21]

Answer:

2

Step-by-step explanation:

3 0
3 years ago
Based on the data which company most likely has the longest average commute time per employee?
Natalija [7]
Company 1 average= 24.4 minutes 
company 2 average= 19.8 minutes
company 3 average= 15 minutes 
company 4 average= 19 minutes

The answer is 24.4 or company one
3 0
3 years ago
According to a study done by a university​ student, the probability a randomly selected individual will not cover his or her mou
VikaD [51]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

 P(X = 8) =  0.0037

b

 P(X <  5) =  0.805

c

 P(X > 6) =  0.0206

I would be surprised because the value is very small , less the 0.05

Step-by-step explanation:

From the question we are told that

The probability a randomly selected individual will not cover his or her mouth when sneezing is p = 0.267

Generally data collected from this study follows  binomial  distribution because the number of trials is  finite , there are only two outcomes, (covering  , and  not covering mouth when sneezing ) , the trial are independent

Hence for a randomly selected variable  X we have that  

   X \ \ \~ \ \ { B ( p , n )}

The probability distribution function for binomial  distribution is  

    P(X = x ) =  ^nC_x *  p^x *  (1 -p) ^{n-x}

Considering question a

Generally the  the probability that among 12 randomly observed individuals exactly 8 do not cover their mouth when​ sneezing is mathematically represented as

     P(X = 8) =  ^{12} C_8 *  (0.267)^8 *  (1- 0.267)^{12-8}

Here C denotes  combination

So

     P(X = 8) =  495  *  0.000025828 * 0.28867947

    P(X = 8) =  0.0037

Considering question b

Generally the probability that among 12 randomly observed individuals fewer than 5 do not cover their mouth when​ sneezing is mathematically represented as

     P(X <  5 ) =[P(X = 0 ) + \cdots + P(X = 4)]

=>   P(X <  5 ) =[ ^{12} C_0 *  (0.267)^0 *  (1- 0.267)^{12-0} + \cdots +  ^{12} C_4 *  (0.267)^4 *  (1- 0.267)^{12-4} ]

=> P(X <  5 )  =  0.02406 +  0.10516 + 0.21067 + 0.25580 + 0.20964

=>  P(X <  5) =  0.805

Considering question c

Generally the probability that fewer than half(6) covered their mouth when​ sneezing(i.e the probability the greater than half do not cover their mouth when sneezing) is mathematically represented as

      P(X > 6) =  1 - p(X \le  6)

=>    P(X > 6) = 1 - [P(X = 0) + \cdots + P(X =6)]

=>    P(X > 6)=1 - [^{12} C_0 *  (0.267)^0 *  (1- 0.267)^{12-0}+ \cdots + ^{12} C_4 *  (0.267)^6 *  (1- 0.267)^{12-6} ]

=>    P(X > 6)= 1 - [0.02406 + \cdots + 0.0519 ]  

=>    P(X > 6) =  0.0206

I would be surprised because the value is very small , less the 0.05

8 0
3 years ago
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