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Mrrafil [7]
3 years ago
5

-3x > 24 on a number line

Mathematics
1 answer:
lianna [129]3 years ago
4 0

\text {Hey! Here is the number line you requested}

\text {This equation will also equal...}

x

\text {How do you get that answer?} \text {Simple! You just divide both sides by 3}

-3x/3>24/-3

\text {Final Answer}

x

\text {Best of Luck!}

\text {-LimitedX}

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I need some help finding the volume of a sphere with a diameter of 18.6cm. The website says the answer I got from my calculator
Dahasolnce [82]

Answer:

Volume, V = 3367.6 cm³

Step-by-step explanation:

Given the following data;

Diameter = 18.6 cm

Radius = diameter/2 = 18.6/2 = 9.3 cm

Pie, π = 3.14

To find the volume of the sphere;

Mathematically, the volume of a sphere is given by the formula;

Volume \; of \; a \; sphere = \frac {4}{3} \pi r^{3}

Where;

r is the radius of the sphere.

Substituting into the formula, we have;

V = \frac {4}{3} * 3.14 * 9.3^{3}

V =  4.186666667 * 804.357

Volume, V = 3367.6 cm³

6 0
3 years ago
-2 (x + 4) – (3x – 8)<br> What it equals
Alja [10]

Answer:

-2x+-8-3x-8

Step-by-step explanation:

-2 times x =-2x  

-2 times 4=-8

8 0
3 years ago
Read 2 more answers
Point K is on line segment JL. Given KL = 6 and JK = 14, determine the length<br> JL.
mestny [16]
JL= 20

JK+KL= JL
14+6= 20
8 0
3 years ago
How do you simplify this?
lions [1.4K]

Answer:

sinA(cosA+sinA)*cosA

Step-by-step explanation:

(sinA(2cosA+sinA-cosA))/secA

sinA(cosA+sinA)/secA

sinA(cosA+sinA)*cosA

8 0
3 years ago
The prior probabilities for events A1 and A2 are P(A1) = 0.35 and P(A2) = 0.50. It is also known that P(A1 ∩ A2) = 0. Suppose P(
forsale [732]

Answer:

Step-by-step explanation:

Hello!

Given the probabilities:

P(A₁)= 0.35

P(A₂)= 0.50

P(A₁∩A₂)= 0

P(BIA₁)= 0.20

P(BIA₂)= 0.05

a)

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

b)

Considering that P(BIA)= \frac{P(AnB)}{P(A)} you can clear the intersection from the formula P(AnB)= P(B/A)*P(A) and apply it for the given events:

P(A_1nB)= P(B/A_1) * P(A_1)= 0.20*0.35= 0.07

P(A_2nB)= P(B/A_2)*P(A_2)= 0.05*0.50= 0.025

c)

The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:

P(B)= (A₁∩B) + P(A₂∩B)=  0.07 + 0.025= 0.095

d)

The Bayes' theorem states that:

P(Ai/B)= \frac{P(B/Ai)*P(A)}{P(B)}

Then:

P(A_1/B)= \frac{P(B/A_1)*P(A_1)}{P(B)}= \frac{0.20*0.35}{0.095}= 0.737 = 0.74

P(A_2/B)= \frac{P(B/A_2)*P(A_2)}{P(B)} = \frac{0.05*0.50}{0.095} = 0.26

I hope it helps!

5 0
3 years ago
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