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Diano4ka-milaya [45]

3 years ago

10

A waether forecast predicts a 30% chance of rain for each of the next 3 days. Describe a way to stimulate the chance that it wil

l rain the next three days

1 answer:

DerKrebs [107]3 years ago

5 0

Using a binomial distribution considering there's a 30% chance it will rain on any of the three days:

<span>The probability of it raining on 0 days is (1)(0.7)(0.7)(0.7) = 34.3%. </span>
<span>The probability of it raining on 1 day is (3)(0.3)(0.7)(0.7) = 44.1%. </span>
<span>The probability of it raining on 2 days is (3)(0.3)(0.3)(0.7) = 18.9%. </span>
<span>The probability of it raining on 3 days is (1)(0.3)(0.3)(0.3) = 2.7%. </span>

<span>There's a 65.7% chance that it will rain at least once over the three-day period.</span>

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3 years ago

Suppose that the quarterly sales levels among health care information systems companies are approximately normally distributed w

jok3333 [9.3K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The sales level that is the cut-off between quarters that are considers as "failure" and those that are not is evaluated as

$y = 6.3 \ million \ dollars$

Step-by-step explanation:

From the question we are told that the

Mean is $\mu = 8 \ million \ dollars$

Standard deviation is $\sigma = 1.3 \ million \ dollars$

Let Y be the random variable that denotes the sales made quarterly among health care information system

The normal distribution for this data is mathematically represented as

$Y$ ~ N $(\mu = 8 , \sigma^2 = 1.69)$

From the question we are told that a quarter is consider a failure by the company if the sales level that quarter in the bottom 10% of the of all quarterly sales

So

The the Probability of obtaining quarterly sales level that is equal to 10% of all quarterly sale (It is not below 10%) is mathematically represented as

$P[Y < y ] = 0.10$

This can be represented as a normal distribution in this manner

$P [ \frac{Y- \mu}{\sigma } < \frac{y -\mu}{\sigma} ] = 0.10$

Where $\frac{Y - \mu }{\sigma } = Z$ ~ $N(0 , 1)$

{This mean that this equal to the normal distribution between 0, 1 which is the generally range of every probability }

Therefor we have

$P[Z < \frac{y- 8}{1.3} ]$

The cumulative distribution function for a normal distribution of y is mathematically represented as

$\phi [\frac{y- 8}{1.3} ] = 0.10$

This is because a cumulative distribution function of a random value Y or a distribution of Y evaluated at y is the probability that Y will take will take a value that is less or equal to y

$\frac{y- 8}{1.3} = \phi ^{-1}(0.10)$

Calculating the inverse of the cumulative distribution function value of 0.10 which is negative the the critical value(Critical values determine what probability a particular variable will have when a sampling distribution is normal or close to normal.) of 0.01

$\phi ^{-1}(0.10) =- [1 - \frac{0.10}{2}]$

$= - 1.28155$

$\frac{y - 8}{1.3} = - 1.2816$

$=> y = 8 - (1.2816)(1.3)$

$y = 6.3 \ million \ dollars$

5 0

3 years ago

-2x + y = 0<br> x - y = -3

miv72 [106K]

Answer:

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Step-by-step explanation:

You can solve this simultaneous equation by elimination method :

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$(1) + (2)$

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$x = 3$

$substitute \: x = 3 \: into \: (2)$

$(2)⇒$

$3 - y = - 3$

$- y = - 6$

$y = 6$

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3 years ago

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dem82 [27]

4*20%
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AB
It is indicated in the figure and is valid:
m AB = 49
m ABC
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Angle resulting from the sum of the two angles shown in the figure:
m BAC = 107 + 49 = 156

m ACB
Angle resulting from the following subtraction:
m ACB = 360-49 = 311

5 0

3 years ago

Read 2 more answers