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3 years ago

Find the 11th term of the arithmetic sequence graphed below. 6 3 -3 -6 -9 -12 -15 -18

Mathematics

1 answer:

Yakvenalex [24]3 years ago

8 0

-57 if you keep following the pattern

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olivia goes to egypt and saw a model of a pyramid. She enlarged it using a scale factor of 1.2. What will be the perimeter and a

Nana76 [90]

The perimeter will be the perimeter of the real thing divided by 1.2 .

The area will be the area of the real thing divided by (1.2)² .

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3 years ago

7%of what length is 200 feet?

Aneli [31]

Step-by-step explanation:

Let the length be x

7% of x = 200ft

7/100 × x = 200

7x/100 = 200

7x = 200 × 100

7x = 20000

x = 20000/7

x = 2,857.14

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3 years ago

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P is the product of 7 to the sum of 3 and 9

Colt1911 [192]

The equation: P×7+3+9

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4 years ago

What is the negative inverse of -0.3

TEA [102]

$n\to negative\ of\ n=-n\\\\n\to inverse\ of\ n=\dfrac{1}{n}\\\\n\to negative\ inverse\ of\ a=-\dfrac{1}{a}\\\\-0.3=-\dfrac{3}{10}\\\\negative\ inverse:\\\\-\left(-\dfrac{10}{3}\right)=\dfrac{10}{3}$

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3 years ago

At an outpatient mental health clinic, appointment cancellations occur at a mean rate of 1.5 per day on a typical Wednesday. Let

Kamila [148]

Answer:

a) Cancellations are independent and similar to arrivals.

b) 22.31% probability that no cancellations will occur on a particular Wednesday

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.

Mean rate of 1.5 per day on a typical Wednesday.

This means that $\mu = 1.5$

(a) Justify the use of the Poisson model.

Each wednesday is independent of each other, and each wednesday has the same mean number of cancellations.

So the answer is:

Cancellations are independent and similar to arrivals.

(b) What is the probability that no cancellations will occur on a particular Wednesday

This is P(X = 0).

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

$P(X = 0) = \frac{e^{-1.5}*(1.5)^{0}}{(0)!} = 0.2231$

22.31% probability that no cancellations will occur on a particular Wednesday

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4 years ago

Find the 11th term of the arithmetic sequence graphed below. 6 3 -3 -6 -9 -12 -15 -18​

Find the 11th term of the arithmetic sequence graphed below. 6 3 -3 -6 -9 -12 -15 -18