Tape Diagrams and Writing Equations

Showing powers of tens using exponents

Triss [41]

10^1. 10^2. 10^3. 10^4. 10^5. Etc

7 0

3 years ago

I NEED SOMEONE TO GIVE ME THE CORRECT ANSWERS TO THIS

kipiarov [429]

1. x-5
2. 12x + 5
3. x/5
4. 3x - (x + 5)
hope i helped :)

7 0

3 years ago

EASY MATH THAT I DON,T GET PLZ HELP QUICKLY<br>

jolli1 [7]

Answer:

Terms: 4

Degree: 3

Leading coeff: 7

Step-by-step explanation:

a term is either a single number or variable, or numbers and variables multiplied together. Terms are separated by + or − signs, or sometimes by division. 7x³+5x²-3x+1 would have 4 terms: 7x³, 5x², -3x, and 1

You already have the degree which is the highest power in a polynomial

The leading coefficent is the coefficient in the term with the highest power, in this case being 7.

hope this helps!

6 0

3 years ago

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

A41.0m guy wire is attached to the top of a 32.8 m antenna and to a point on the ground. How far is the point on the ground from

Setler79 [48]

Answer:

Step-by-step explanation:

A right angle triangle is formed.

The length of the guy wire represents the hypotenuse of the right angle triangle.

The height of the antenna represents the opposite side of the right angle triangle.

The distance, h from base of the antenna to the point on the ground to which the antenna is attached represents the adjacent side of the triangle.

To determine h, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

Therefore,

41² = 32.8² + h²

1681 = 1075.84 + h²

h² = 1681 - 1075.84 = 605.16

h = √605.16

h = 24.6 m

To determine the angle θ that the wire makes with the ground, we would apply the the cosine trigonometric ratio.

Cos θ = adjacent side/hypotenuse. Therefore,

Cos θ = 24.6/41 = 0.6

θ = Cos^-1(0.6)

θ = 53.1°

7 0

3 years ago