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3 years ago

Help plzzzzzz 10 point question

Mathematics

2 answers:

Rus_ich [418]3 years ago

5 0

Danville,Keflavik,Somerville,Ulsan,,Smithfield

xeze [42]3 years ago

3 0

Answer:

-22,-13,-6,15,18

Step-by-step explanation:

with negative numbers if it is higher with a - then it is acctually lower, and when it is positive then higher means higher. Think of it as how far away the number is from zero. so you count which is the farthest negative to which is the farthest positive.

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Weary of the low turnout in student elections, a college administration decides to choose an SRS of three students to form an ad

-Dominant- [34]

Answer:

$P(ABC) = 0.110592$

$P(ABC^c) = 0.119808$

$P(AB^cC) = 0.119808$

$P(A^cBC) = 0.119808$

$P(AB^cC^c) = 0.129792$

$P(A^cBC^c) = 0.129792$

$P(A^cB^cC) = 0.129792$

$P(A^cB^cC^c) = 0.140608$

Step-by-step explanation:

Given

$P(A) = P(B) = P(C) = 48\%$

Convert the probability to decimal

$P(A) = P(B) = P(C) = 0.48$

Solving (a): P(ABC)

This is calculated as:

$P(ABC) = P(A) * P(B) * P(C)$

This gives:

$P(ABC) = 0.48*0.48*0.48$

$P(ABC) = 0.110592$

Solving (b): $P(ABC^c)$

This is calculated as:

$P(ABC^c) = P(A) * P(B) * P(C^c)$

In probability:

$P(C^c) = 1 - P(C)$

So, we have:

$P(ABC^c) = P(A) * P(B) * (1 - P(C))$

$P(ABC^c) = 0.48 * 0.48 * (1 - 0.48)$

$P(ABC^c) = 0.48 * 0.48 * 0.52$

$P(ABC^c) = 0.119808$

Solving (c): $P(AB^cC)$

This is calculated as:

$P(AB^cC) = P(A) * P(B^c) * P(C)$

$P(AB^cC) = P(A) * [1 - P(B)] * P(C)$

$P(AB^cC) = 0.48 * (1 - 0.48)* 0.48$

$P(AB^cC) = 0.48 * 0.52* 0.48$

$P(AB^cC) = 0.119808$

Solving (d): $P(A^cBC)$

This is calculated as:

$P(A^cBC) = P(A^c) * P(B) * P(C)$

$P(A^cBC) = [1-P(A)] *P(B) * P(C)$

$P(A^cBC) = (1 - 0.48)* 0.48 * 0.48$

$P(A^cBC) = 0.52* 0.48 * 0.48$

$P(A^cBC) = 0.119808$

Solving (e): $P(AB^cC^c)$

This is calculated as:

$P(AB^cC^c) = P(A) * P(B^c) * P(C^c)$

$P(AB^cC^c) = P(A) * [1-P(B)] * [1-P(C)]$

$P(AB^cC^c) = 0.48 * [1-0.48] * [1-0.48]$

$P(AB^cC^c) = 0.48 * 0.52*0.52$

$P(AB^cC^c) = 0.129792$

Solving (f): $P(A^cBC^c)$

This is calculated as:

$P(A^cBC^c) = P(A^c) * P(B) * P(C^c)$

$P(A^cBC^c) = [1-P(A)] * P(B) * [1-P(C)]$

$P(A^cBC^c) = [1-0.48] * 0.48 * [1-0.48]$

$P(A^cBC^c) = 0.52 * 0.48 * 0.52$

$P(A^cBC^c) = 0.129792$

Solving (g): $P(A^cB^cC)$

This is calculated as:

$P(A^cB^cC) = P(A^c) * P(B^c) * P(C)$

$P(A^cB^cC) = [1-P(A)] * [1-P(B)] * P(C)$

$P(A^cB^cC) = [1-0.48] * [1-0.48] * 0.48$

$P(A^cB^cC) = 0.52 * 0.52 * 0.48$

$P(A^cB^cC) = 0.129792$

Solving (h): $P(A^cB^cC^c)$

This is calculated as:

$P(A^cB^cC^c) = P(A^c) * P(B^c) * P(C^c)$

$P(A^cB^cC^c) = [1-P(A)] * [1-P(B)] * [1-P(C)]$

$P(A^cB^cC^c) = [1-0.48] * [1-0.48] * [1-0.48]$

$P(A^cB^cC^c) = 0.52*0.52*0.52$

$P(A^cB^cC^c) = 0.140608$

5 0

3 years ago

Solve 2x2 + 20x = −38.

cestrela7 [59]

2x^2+20x=-38 divide both sides by 2

x^2+10x=-19, halve the linear coefficient, square it, than add that value to both sides of the equation, in this case add (10/2)^2=25...

x^2+10x+25=6 now the left side is a perfect square

(x+5)^2=6 take the square root of both sides

x+5=±√6 subtract 5 from both sides

x=-5±√6 (so answer c.)

4 0

3 years ago

Are Functions 1 and 2 the same? Explain. Function 1: y = 3(x + 5) Function 2: y equals three times x, plus 5.

Molodets [167]

Answer:

No, the functions are not the same.

Step-by-step explanation:

For function one you have to distribute the three to the x and the five which would give you a function of 3x+15.

This will result in different y-intercepts function one's why being 15 and function's two y-intercept being 5. Though the will have the same slope they are two different functions.

7 0

3 years ago

Find The Area Of This Shape

spin [16.1K]

14 because you count the height and how much the base is from the farthest point left to right and multiply h=2 b=7 therefor 2x7 is 14

7 0

3 years ago

Which values are solutions to the inequality below? √x ≤ 5

Romashka [77]

$\sqrt{x} \leq 5$

Find the real region for $\sqrt{x}$

$\sqrt{x}$ is real for $x \geq 0$

$= x \geq 0$

$\sqrt{x} \leq 5$

Square both sides:

$( \sqrt{x} )^2 \leq 5^2$

$5^2 = 5*5 =25$

Refine:

$x \leq 25$

Answer D

hope this helps!

5 0

3 years ago