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3 years ago

A series of tile patterns is shown below. Consider the function that represents the number of white

Mathematics

1 answer:

atroni [7]3 years ago

3 0

Answer:

Step-by-step explanation:

first you add them and then you subtract and you will get your answer

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Helpppp please I’ll name brainliest

Dominik [7]

Answer: Commutative Property of.

Addition

Step-by-step explanation:

3 0

3 years ago

The number of visitors to a certain Web site triples every month. The number of visitors is modeled by the expression 8100 times

olasank [31]

Suppose we have had S visitors in n months since EIS (Entry Into Service) of the site, then :
S(n) = 8100 + 8100 * 3 + 8100 * 3^2 + 8100 * 3^3 + ... +8100 * 3^n 
S(n) = 8100 * [ 1 + 3 + 3^2 + 3^3 + ... + 3^n ] 
and we know that : 
1 + 3 + 3^2 + 3^3 + ... + 3^n = (3^(n+1) - 1)/(3 - 1) 
= (3^(n+1) - 1)/2 
therefore 
S(n) = 8100 * (3^(n+1) - 1)/2 
finally : 
S(n) = 4050 * (3^(n+1) - 1) 
and this allows you to obtain the month n, if you know S(n) : 
3^(n+1) = S(n) /4050 + 1 
therefore : 

n = ln[S(n) /4050 + 1] /ln3 - 1 

hope it' ll help !!

6 0

2 years ago

certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 80% of the time if the pe

Alekssandra [29.7K]

Answer:

The probability that a person has the virus given that they have tested positive is P(A|B)=0.032.

The probability that a person does not have the virus given that they have tested negative is P(A'|B')=0.999.

Step-by-step explanation:

a) The probability of B: "the person tests positive" is the sum of:

- The probability that a person has the virus and the test is positive.

$P(A) \cdot P(B|A)$

- The probability that a person don't have the virus but the test is positive.

$P(not \,A) \cdot P(B|not\,A)$

Then, the probability of B: "the person tests positive" is

$P(A|B)=\frac{P(B|A)}{P(B)}=\frac{P(A)\cdot P(B|A)}{P(A)\cdot P(B|A)+P(A')\cdot P(B|A')}$

We have

$P(A)=1/300=0.0033\\\\P(A')=1-P(A)=0.9967\\\\P(B|A)=0.8\\\\P(B|A')=0.08$

Then

$P(A|B)=\frac{P(A)\cdot P(B|A)}{P(A)\cdot P(B|A)+P(A')\cdot P(B|A')}\\\\P(A|B)=\frac{0.0033*0.8}{0.0033*0.8+0.9967*0.08}= \frac{0.00264}{0.00264+0.07974}=\frac{0.00264}{0.08238}= 0.032$

The probability that a person has the virus given that they have tested positive is P(A|B)=0.032.

b) The probability P(A'|B') is

$P(A'|B')=\frac{P(B'|A')\cdot P(A')}{P(B'|A')\cdot P(A')+P(B'|A)\cdot P(A)}$

We have

$P(A)=0.0033\\\\P(A')=1-P(A)=0.9967\\\\P(B'|A)=1-0.8=0.2\\\\P(B'|A')=1-0.08=0.92$

Then

$P(A'|B')=\frac{P(B'|A')\cdot P(A')}{P(B'|A')\cdot P(A')+P(B'|A)\cdot P(A)}=\frac{0.92*0.9967}{0.92*0.9967+0.2*0.0033}= \frac{0.917}{0.917+0.00066}= 0.999$

The probability that a person does not have the virus given that they have tested negative is P(A'|B')=0.999.

5 0

3 years ago

Cody has a strip of paper 10 centimeters long he cut it into three pieces one piece is 4 centimeters long the second piece is 3

jolli1 [7]

The starting length of the strip is 10cm
he cuts the first piece (4cm):10-4=6
that means there's 6cm left
he cuts the second piece (3cm):6-3=3
that means there's 3cm left
the answer is the third piece is 3cm long

6 0

2 years ago

Find the equation for the tangent line of f(x)=−3x2−7x+3 at x=3.

kirza4 [7]

Hello :
the equation for the tangent line at point : A(x' , y' ) is :
y - y' = f'(x') (x-x')........f'(x') the slope
in this exercice :
f(x) = -3x²-7x+3 and : x' = 3
y' = f(3) = -3(3)²-7(3) +3 =- 45
but : f'(x) = -6x-7 ...( derivate of : f(x) )
f'(3) = -6(3) -7 = - 25

the equation for the tangent line is : y - (-45) = -25(x-3)
y = -25x+30

6 0

3 years ago