Express the number in scientific notation. 3.26

Please help me please

Levart [38]

Answer:

Alternate Interior Angles Theory

Step-by-step explanation:

Any number inside the angle is known as an Alternate Interior Angle.

7 0

2 years ago

A collector of scale model of ships wants to know the length of a 1/69 scale model of a ship that was 207 feet in length. Find t

Tresset [83]

I think 3 because if u divide 207 by 69 you get 3

4 0

3 years ago

How do you do this problem

garri49 [273]

Answer:

110

Step-by-step explanation:

8x10=80

Area of triangle= 1/2 base times height

3x10=30

80+30=110

7 0

3 years ago

Find parametric equations through point P=(2,2,7) in the direction of the vector v = (44, 14, -20)?

ycow [4]

9514 1404 393

Answer:

(x, y, z) = (2+44t, 2+14t, 7-20t)

Step-by-step explanation:

One way to write parametric equations for line L is ...

L = P + t·<em>v</em>

where P is the given point and <em>v</em> is the given direction vector. Using that form, we have ...

(x, y, z) = (2+44t, 2+14t, 7-20t)

__

If you like, you can remove a common factor of 2 from the coefficients of t.

(x, y, z) = (2+22t, 2+7t, 7-10t)

5 0

2 years ago

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

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 sucesses

e = 2.71828 is the Euler number

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

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

8 0

3 years ago

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