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3 years ago

What is the result of dividing -8 by a number less than -2?

Mathematics

1 answer:

DedPeter [7]3 years ago

4 0

You get a positive number.

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8. Solve the following proportion (1 point)

Arturiano [62]

Answer:

97.5

Step-by-step explanation:

Cross multiply

2 * x = 2x

15 * 13 = 195

2x = 195

2x/2 = 195/2

x = 97.5

[RevyBreeze]

6 0

2 years ago

Read 2 more answers

Which of the following is NOT true about P-values in hypothesis testing? Choose the correct answer below. A. If the P is high

pashok25 [27]

Answer:

Step-by-step explanation:

In statistical hypothesis testing we use p values. For any test we select significance level alpha as 5% or 1% or 10%

Whenever p < alpha we reject our null hypothesis. Otherwise we accept our null hypothesis.

Hence p should be high to accept null hypothesis

Option A is incorrect

B is correct

C is wrong as p is the probability

D False

Option B is right.

7 0

3 years ago

Magdalena Figueroa's covered medical care bills for a recent illness were $8,440.00. Her major medical policy had a coinsurance

devlian [24]

Medical bill = $8,440
Deductible = $400
Coinsurance plan = Medical policy 80% of all expenses above $400

Amount to pay = 400+ (8440-400)*20/100 = 400+1608 = $2,008

8 0

3 years ago

The zeros of the functions f(x) = 3x^2 - 3x - 6

Butoxors [25]

The answer is: x = 2 ∨ x = -1

7 0

2 years ago

The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of si

OverLord2011 [107]

Answer:

a) P(x=3)=0.089

b) P(x≥3)=0.938

c) 1.5 arrivals

Step-by-step explanation:

Let t be the time (in hours), then random variable X is the number of people arriving for treatment at an emergency room.

The variable X is modeled by a Poisson process with a rate parameter of λ=6.

The probability of exactly k arrivals in a particular hour can be written as:

$P(x=k)=\lambda^{k} \cdot e^{-\lambda}/k!\\\\P(x=k)=6^k\cdot e^{-6}/k!$

a) The probability that exactly 3 arrivals occur during a particular hour is:

$P(x=3)=6^{3} \cdot e^{-6}/3!=216*0.0025/6=0.089\\\\$

b) The probability that <em>at least</em> 3 people arrive during a particular hour is:

$P(x\geq3)=1-[P(x=0)+P(x=1)+P(x=2)]\\\\\\P(0)=6^{0} \cdot e^{-6}/0!=1*0.0025/1=0.002\\\\P(1)=6^{1} \cdot e^{-6}/1!=6*0.0025/1=0.015\\\\P(2)=6^{2} \cdot e^{-6}/2!=36*0.0025/2=0.045\\\\\\P(x\geq3)=1-[0.002+0.015+0.045]=1-0.062=0.938$

c) In this case, t=0.25, so we recalculate the parameter as:

$\lambda =r\cdot t=6\;h^{-1}\cdot 0.25 h=1.5$

The expected value for a Poisson distribution is equal to its parameter λ, so in this case we expect 1.5 arrivals in a period of 15 minutes.

$E(x)=\lambda=1.5$

3 0

3 years ago