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2 years ago

PLEASE HELP 2 QUESTIONS 45 POINTS!!!!!!!!!

Mathematics

1 answer:

Effectus [21]2 years ago

5 0

Answer:

1. -8x² - 16x - 6 a = -8, b = -16, c = -6

1. 4x² - 1 a = 4, b = 0, c = -1

Step-by-step explanation:

use the 'foil' method:

(4x + 2)(-2x - 3)

F: -8x² O: -12x I: -4x L: -6

-8x² - 16x - 6

(2x + 1)(2x - 1)

F: 4x² O: -2x I: 2x L: -1

4x² - 1

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A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 14 pa

UkoKoshka [18]

Answer:

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

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 interval.

Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.

Each minute has 60 seconds, so $\mu = \frac{14}{60} = 0.2333$

Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = \frac{e^{-0.2333}*(0.2333)^{0}}{(0)!} = 0.7919$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7919 = 0.2081$

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

8 0

3 years ago

Answer and show work #36

Artyom0805 [142]

I dont understant that

8 0

3 years ago

Please help me Aspa Plaease

Natasha_Volkova [10]

She need 3 bags since the area of the triangle is base x height /2
Area 20*6=120
120/2=60
60/20= 3 since each bag covers 20m^2

7 0

2 years ago

Days before a presidential election, a nationwide random sample of registered voters was taken. Based on this random sample, i

erica [24]

Answer:

We are 95% confident that the percentage of registered voters in the nation planning on voting for Robert Smith is between 49% and 55%.

Step-by-step explanation:

Given that :

Margin of Error = ±3%

Sample Proportion = 52%

Confidence level = 95%

The 95% confidence interval is :

Sample proportion ± margin of error

52% ± 3%

Lower boundary = 52% - 3% = 49%

Upper boundary = 52% + 3% = 55%

The interpretation is that at a given confidence level ; the popukation proportion based on the sample proportion and margin of error is in the confidence interval.

7 0

2 years ago

In a random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand. C

horrorfan [7]

Answer:

90% confidence interval for the proportion of fans who bought food from the concession stand

(0.5603,0.6529)

Step-by-step explanation:

Step(i):-

Given sample size 'n' =300

Given data random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand.

Given sample proportion

$p^{-} = \frac{x}{n} = \frac{182}{300} =0.606$

level of significance = 90% or 0.10

Z₀.₁₀ = 1.645

90% confidence interval for the proportion is determined by

$(p^{-} - Z_{0.10}\sqrt{\frac{p(1-p)}{n} } , p^{-} +Z_{0.10}\sqrt{\frac{p(1-p)}{n} })$

$(0.6066 - 1.645\sqrt{\frac{0.6066(1-0.6066)}{300} } ,0.6066+1.645\sqrt{\frac{0.6066(1-0.6066)}{300} })$

(0.6066 - 0.0463 ,0.6066 + 0.0463)

(0.5603,0.6529)

final answer:-

90% confidence interval for the proportion of fans who bought food from the concession stand

(0.5603,0.6529)

5 0

2 years ago