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3 years ago

10

Does anybody know the answer to this question Im having lots of problems

2 answers:

Dvinal [7]3 years ago

7 0

<h3>Answer: 3^9</h3>

This is the same as writing $3^9$

====================================================

How to get that answer:

The lowest height is 3^8, and the highest point is 3 times that value.

We can think of that second "3" as really 3^1. This is because x^1 = x for any real number.

Multiply 3^8 and 3^1 using the rule that a^b*a^c = a^(b+c). We add the exponents together.

Therefore, 3^8*3^1 = 3^(8+1) = 3^9

Side notes:

3^8 = 6,561
3^9 = 19,683

Lunna [17]3 years ago

3 0

Answer:

3^9

Step-by-step explanation:

3 * 3^8

3^1 * 3^8

3^1+8

= 3^9

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Help me yall!!......

Ugo [173]

This then allows us to see exactly how and where the subtended angle θ of a sector will fit into the sector formulas. Now we can replace the "once around" angle (that is, the 2π) for an entire circle with the measure of a sector's subtended angle θ, and this will give us the formulas for the area and arc length of that sector

8 0

2 years ago

Consider a Poisson distribution with μ = 6.

bearhunter [10]

Answer:

a) $P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b) f(2) = 0.04462

c) f(1) = 0.01487

d) $P(X \geq 2) = 0.93803$

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

In this question:

$\mu = 6$

a. Write the appropriate Poisson probability function.

Considering $\mu = 6$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b. Compute f (2).

This is P(X = 2). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

Then

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803$

So

$P(X \geq 2) = 0.93803$

5 0

2 years ago

|x - 3|<5<br><br> Pls solve this you need to rewrite it without modulus sign and explain as well

Rama09 [41]

Answer:

-2< x< 8

Step-by-step explanation:

|x-3|<5 <=> -5<x-3<5 <=> -2<x<8

..

3 0

2 years ago

Read 2 more answers

-4/5x - 20/7x = 663/56 - 1/2x

svetlana [45]

Answer:

I believe the answer to this question is: X= -3 3/4

7 0

3 years ago

Please help me <br> Please answer this

statuscvo [17]

Triangle ABC has vertices A, B and C at points (-3,4), (-3,0) and (-1,3), respectively.

1. First transformation is rotation by 90° anti-clockwise about the origin. This transformation has a rule:

(x,y)→(-y,x).

Then

A(-3,4)→A''(-4,-3);
B(-3,0)→B''(0,-3);
C(-1,3)→C''(-3,-1).

2. Second transformation is translation 2 units to the right with a rule

(x,y)→(x+2,y).

Then,

A''(-4,-3)→A'(-2,-3);
B''(0,-3)→B'(2,-3);
C''(-3,-1)→C'(-1,-1).

Answer: First transformation is rotation by 90° anti-clockwise about the origin, second transformation is translation 2 units to the right.

5 0

3 years ago