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-Dominant- [34]
3 years ago
13

Why is it sometimes best to make a phone call or personal contact when conveying sensitive information?

Mathematics
1 answer:
kupik [55]3 years ago
3 0
It would be best sometimes to <span>make a phone call or personal contact when conveying sensitive information because in this way you can make sure that the information is conveyed directly to the person. Hope this answers the question. Have a nice day.</span>
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HELP PLS Match each graph with the logarithmic function it represents
SIZIF [17.4K]

Each graph has been matched with the logarithmic function it represents as follows:

  • f(x) = 3 - 4In (x-2) = graph 3.
  • f(x) = 3 - Inx = graph 1.
  • f(x) = In(x + 1) = graph 4.
  • f(x) = 2In(x + 3) = graph 2.

<h3>What is a function?</h3>

A function can be defined as a mathematical expression which is used to define and represent the relationship that exists between two or more variables.

<h3>The types of function.</h3>

In Mathematics, there are different types of functions and these include the following;

  • Periodic function
  • Inverse function
  • Modulus function
  • Signum function
  • Piece-wise defined function.
  • Logarithm function

<h3>What is a logarithm function?</h3>

A logarithm function can be defined as a type of function that represents the inverse of an exponential function. Mathematically, a logarithm function is written as follows:

y = logₐₓ

In this exercise, you're required to match each graph with the logarithmic function it represents as shown in the image attached below:

  • f(x) = 3 - 4In (x-2) = graph 3.
  • f(x) = 3 - Inx = graph 1.
  • f(x) = In(x + 1) = graph 4.
  • f(x) = 2In(x + 3) = graph 2.

Read more on logarithm function here: brainly.com/question/13473114

#SPJ1

3 0
2 years ago
Someone please explain and answer it. There's a picture with one problem. Thank you!
Tatiana [17]
Site vandal. Do not answer.
6 0
3 years ago
The probability that it will rain on Thursday is 67%. What is the probability that it won’t rain on Thursday? Express your answe
kondaur [170]

Answer:

33%

Step-by-step explanation:

Since the probability of rain on Thursday is 67%, the probability of no rain on Thursday is 100% - 67% = 33%.

4 0
2 years ago
Can someone that's very good at math please help me?
gtnhenbr [62]

Answer:

35 :

t = 6.25 years

(about 6 years 3 months)

Equation:

t = (1/r)(A/P - 1)

Calculation:

First, converting R percent to r a decimal

r = R/100 = 4%/100 = 0.04 per year,

then, solving our equation

t = (1/0.04)((2500/2000) - 1) = 6.25

t = 6.25 years

The time required to get a total amount, principal plus interest, of $2,500.00 from simple interest on a principal of $2,000.00 at an interest rate of 4% per year is 6.25 years (about 6 years 3 months).

36:

The two distances are the same (out and back), so set them equal.

That is done by having a (rate)(time) equal a (rate)(time).

One time is “x” and the other is “4.8-x.”

One rate is 460 and the other is 500.

460 x = 500 (4.8 -x)

460 x = 2400 - 500x

900 x = 2400

x = 2.5 hours for the slower plane.

4.8- x = 2.3 hours for the faster plane.

8 0
3 years ago
A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute
nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

P(k\geq8)=1-P(k

P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\

P(k

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002

5 0
3 years ago
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