Please help ASAP [IMAGE]

tigry1 [53]

welp thank you for this

7 0

3 years ago

The function f(x) = −x2 − 5x + 50 shows the relationship between the vertical distance of a diver from a pool's surface f(x), in

natta225 [31]

Answer:

<u>The zero is x = 5, and it represents that the diver will enter the pool's surface 5 feet in front of the diving board.</u>

Explanation:

The function is f(x) = - x² - 5x + 50.

Where:

f(x) is the vertical distance of a diver from a pool's surface, and

x is the horizontal distance of a diver from the diving board.

A zero of f(x) represents a point where the diver is at the surface of the pool. There might be two possible zeros.

Solve the equation to determine the zeros:

- x² - 5x + 50 = 0

Multiply both sides by - 1: x² + 5x - 50

Factor: (x + 10) (x - 5) = 0

Use zero product property: x + 10 = 0 or x - 5 = 0

From x + 10 = 0 you get x = - 10

From x - 5 = 0 you get x = 5.

The negative zero is not valid as it would be a point behind the diving board, so the meaningful zero is when x = 5, meaning that the diver will touch the surface of the pool at 5 feet in front of the diving board.

5 0

3 years ago

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A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi

SVETLANKA909090 [29]

Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

$= \frac{20!}{(20-5)! 5!)}$

$= \frac{20*19*18*17*16}{5*4*3*2*1}$

= 15504

b) probabilities of A2, A5, A4

P(A2) = ⁵C₂ / ²⁰C₂

$= \frac{20}{320} = \frac{1}{19}$

P(A5) = ⁸C₅ / ²⁰C₅

$= \frac{56}{15540} = \frac{7}{1938}$

P(A4) = ⁷C₄ / ²⁰C₄

$= \frac{35}{4845} = \frac{7}{969}$

c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1

5 0

3 years ago

Solve the following proportion for x. 3/13 = x/5 Round your answer to the nearest tenth.

anastassius [24]

To solve, isolate the x. Cross multiply

3/13(13)(5) = x/5(5)(13)

3(5) = (13)x

Simplify

13x = 15

Isolate the x. Divide by 13.

13x/13 = 15/13

x = 15/13

x = 1.2

1.2 is your answer

hope this helps

8 0

3 years ago

Find the quotient. 5 1/4 ÷ 1 3/4. *

weeeeeb [17]

Answer:

3

Step-by-step explanation:

私は怠惰なのでグーグルで調べました

5 0

3 years ago

Read 2 more answers