All categories

2 years ago

What is the y-intercept of f(x)=3/5x a. (1, 0) b. (1, 3/5) c. (0, 0) d. (0, 1)

Mathematics

1 answer:

Digiron [165]2 years ago

4 0

Answer
F(0)=0

I can put the work in the comments if you need it :)

You might be interested in

Write your answer using integers, proper fractions, and improper fractions in the simplest form

sattari [20]

Answer:

y = 1/4 x - 7/2

Step-by-step explanation:

x - 4y = 14

-4y = -x + 14

y = -x/(-4) - 14/4

y = 1/4 x - 7/2

6 0

2 years ago

What is x if x/5 =-9

Leona [35]

Answer:

x = -45

Step-by-step explanation:

Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other. Multiply 5 to both sides:

(x/5) * 5 = (-9) * 5

x = -9 * 5

x = -45

x = -45 is your answer.

3 0

3 years ago

Read 2 more answers

Which expression represents the sum (3z−11)+(6−6z) ?

Fudgin [204]

Answer: z = -5/-3

Step-by-step explanation:

(3z−11)+(6−6z)

3z - 11 + 6 - 6z

- 6z + 3z - 11 + 6

- 3z - 5

z = -5/-3

6 0

3 years ago

A new test to detect TB has been designed. It is estimated that 88% of people taking this test have the disease. The test detect

Elodia [21]

Answer:

Correct option: (a) 0.1452

Step-by-step explanation:

The new test designed for detecting TB is being analysed.

Denote the events as follows:

<em>D</em> = a person has the disease

<em>X</em> = the test is positive.

The information provided is:

$P(D)=0.88\\P(X|D)=0.97\\P(X^{c}|D^{c})=0.99$

Compute the probability that a person does not have the disease as follows:

$P(D^{c})=1-P(D)=1-0.88=0.12$

The probability of a person not having the disease is 0.12.

Compute the probability that a randomly selected person is tested negative but does have the disease as follows:

$P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264$

Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:

$P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188$

Compute the probability that a randomly selected person is tested negative as follows:

$P(X^{c})=P(X^{c}\cap D)+P(X^{c}\cap D^{c})$

$=0.0264+0.1188\\=0.1452$

Thus, the probability of the test indicating that the person does not have the disease is 0.1452.

4 0

2 years ago

Which expressions go in which box?

almond37 [142]

To qualify as a polynomial, the expression in question:
* Consists of one or more terms
* Variables are only with positive whole exponents
* No variables in the denominator of any term (the coefficients however, can be fractions.)

In that case the answer is most likely:

3 0

2 years ago