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2 years ago

PLEASE HELP! FUNCTIONS

Mathematics

1 answer:

satela [25.4K]2 years ago

5 0

Answer: 7

Step-by-step explanation:

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Please help me with this problem!

zimovet [89]

2 x $1150 + $1049.34 = $3,349.34.

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2 years ago

Use < , > , = , ≤ , ≥ to complete these statements. 1) if x < y , then y___ x, 2) if x > 5 and 5 > y, then x ___

vodomira [7]

Answer:

1) y > x

2) x > 5 > y

3) x ≤ 4 ( or 4 ≥ x )

4) x = y

Step-by-step explanation:

1) 1. y is greater than x

2. so y > x

2) 1. 5 is less than x and more than y

2. we can write this like x > 5 > y

3. you can just squish them together as a shortcut usually: x > 5 and 5 > y --> x > 5 > y

3) 1. x is for at most, we can say this as x is less than or equal to 4

2. this is written as x ≤ 4 or 4 ≥ x

4) 1. x - y = 0 so x and y must be the same value

2. therefore x = y

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2 years ago

Is 2 pm how many degrees are in the angle between the hour and minute hand

NARA [144]

60° - minute hand is at 90° and hour hand is at 30°. 90-30=60

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2 years ago

What is the constant term in the expression 6x^3y + 8x^2 + 5x + 7? (Input a numeric value only.) (3 points)

Zinaida [17]

The constant term is 7. this is because x and y are unknowns and change depending on the other. As 6x^3y will change depending on what x or y is so will: 8x^2 and 5x therefore 7 is the constant term.

7 0

2 years ago

A new, simple test has been developed to detect a particular type of cancer. The test must be evaluated before it is used. A med

g100num [7]

Answer:

0.3333 = 33.33% probability of a randomly chosen person not having cancer given that the test indicates cancer

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Test indicates cancer.

Event B: Person does not have cancer.

Probability of a test indicating cancer.

98% of 2%(those who have).

1% of 100 - 2 = 98%(those who do not have). So

$P(A) = 0.98*0.02 + 0.01*0.98 = 0.0294$

Probability of a test indicating cancer and person not having.

1% of 98%. So

$P(A \cap B) = 0.01*0.98 = 0.0098$

What is the probability of a randomly chosen person not having cancer given that the test indicates cancer?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0098}{0.0294} = 0.3333$

0.3333 = 33.33% probability of a randomly chosen person not having cancer given that the test indicates cancer

3 0

2 years ago