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2 years ago

A sequence is defined by the equation a1=50 and an=an-1+(-4) what is term a5?

Mathematics

1 answer:

ZanzabumX [31]2 years ago

6 0

Answer:

Step-by-step explanation:

an=an-1+(-4)

a2 = a1 + (-4) = 50 - 4 = 46

a3 = a2 - 4 = 46 - 4 = 42

a4 = a3 - 4 = 42 - 4 = 38

a5 = a4 - 4 = 38 - 4 = 34

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Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe

nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let $D^c$ be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that one person in 10,000 people has a rare genetic disease.

So, $P(D)=\frac{1}{10000}$

Only 0.4% of the people who don't have it test positive.

$P(A|D^c)$ = probability that a person is tested positive given he don't have a disease = 0.004

$P(D^c)=1-\frac{1}{10000}$

Formula: $P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}$

$P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}$

P(D|A)= $\frac{2470}{2471}$ =0.9995

P(D|A)= $0.9995$

A)The probability that someone who tests positive has the disease is 0.9995

(B)

$P(D^c|A^c)$ =probability that someone does not have disease given that he tests negative

$P(A^c|D^c)$ =probability that a person tests negative given that he does not have disease =1-0.004

=0.996

$P(A^c|D)$ =probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: $P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}$

$P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}$

$P(D^c|A^c)=0.99999$

B)The probability that someone who tests negative does not have the disease is 0.99999

8 0

3 years ago

Write a polynomial function f of least degree that has a leading coefficient of 1 and the given zeros -4,-2,5. write the functio

OLga [1]

Answer:

Step-by-step explanation:

This is a third degree polynomial since we have 3 zeros. We find these zeros by factoring the given polynomial. The zeros of a polynomial are where the graph of the function goes through the x-axis (where y = 0). If x = -4, the factor that gives us this value is (x + 4) = 0 and solving that for x, we get x = -4. If x = -2, the factor that gives us that value is (x + 2) = 0 and solving that for x, we get x = -2. Same for the 5. The way we find the polynomial that gave us these zeros is to go backwards from the factors and FOIL them out. That means that we need to find the product of

(x + 4)(x + 2)(x - 5). Do the first 2 terms, then multiply in the third.

$(x+4)(x+2)=x^2+2x+4x+8$ , which simplifies to

$x^2+6x+8$

No we multiply in the final factor of (x - 5):

$(x^2+6x+8)(x-5)=x^3+6x^2+8x-5x^2-30x-40$ which simplifies to

$f(x)=x^3+x^2-22x-40$

If you are aware of the method for factoring higher degree polymomials, which is to use the Rational Root Theorem and synthetic division, you will see that this factors to x = -4, -2, 5. If you know how to use your calculator, you will find the same zeros in your solving polynomials function in your apps.

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3 years ago

There were 100 pizzas sold at Romeo's Pizzeria on Tuesday. A large pizza costs $16 and a small pizza costs $11. If $1,550 worth

Blizzard [7]

Answer: