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3 years ago

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"Jason inherited a piece of land from his great-uncle. Owners in the area claim that there is a 45% chance that the land has oil

. Jason decides to test the land for oil. He buys a kit that claims to have an 80% accuracy rate of indicating oil in the soil. What is the probability that the land has no oil and the test shows that it has oil?
The answers listed are
0.09
0.11
0.44

1 answer:

Bond [772]3 years ago

5 0

Well, if we proceed to plan the problem like this:
<span>A = the land has oil B = the kit is not accurate P(A and B) = P(A) * P(B)
</span>Then we can say that <span>(0.45)(1−0.80)=(0.45)(0.20)=0.09
So the answer will be the first one.
I hope this can help you</span>

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Samuel is running a 3-mile race. He would like to finish the race in under 33 minutes. He has already run for 10.5 minutes. The

Y_Kistochka [10]

For this case we have the following inequality:
$10.5 + x \ \textless \ 33
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From here, we define the variable x.
x: number of minutes remaining to finish the race.
From here, we clear the value of x.
We have then:
$x \ \textless \ 33-10.5
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$x \ \textless \ 22.5
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Therefore, Samuel has less than 22.5 minutes to finish the race in the estimated time.
Answer:
x < 22.5; Sam has fewer than 22.5 minutes left to finish running.

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3 years ago

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What is the solution to the system of equations graphed below?

padilas [110]

Step-by-step explanation:

The solution can be found by identifying the point of intersection of the 2 graphed lines.

As we can see, they intersect each other at the point (2, -3). This is our solution for the equations.

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3 years ago

when vanessa charges her 3ds fully it lasts for 2 hours. if she only charged it 2/8 full how long would it last?

Leto [7]

Answer:

1hour 40min

Step-by-step explanation:

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1hour 40min

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3 years ago

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

yan [13]

Answer:

"The probability that the mean battery life would be greater than 948.8 minutes" is 0.1446.

Step-by-step explanation:

In this case, the quality control expert takes a <em>sample</em> of batteries. From these batteries, we want to find "the probability that the mean battery life would be greater than 948.8 minutes".

Different concepts needed to take into account to solve this question

Sampling Distribution of the Means

For doing this, we need to use the sampling distribution of the means, which results from taking the mean for each possible sample coming from a random variable $\\ x$ . Roughly speaking, each sample will have a different mean, $\\ \overline{x}$ , and the probability distribution for any of these means is called the <em>sampling distribution of the means</em>.

The sampling distribution of the means has a mean that equals the population's mean for the random variable $\\ x$ , i.e., $\\ \mu$ , and its standard deviation is $\\ \frac{\sigma}{\sqrt{n}}$ . We can express this mathematically as:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

Standardized Values for $\\ \overline{x}$

We can standardized the values for $\\ \overline{x}$ using <em>z-scores</em>:

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

This random variable $\\ Z$ follows a <em>standard normal distribution</em>, that is, $\\ Z \sim N(0,1)$ , and it is easier to find probabilities since the values for them are tabulated in the <em>standard normal table</em> (available in any Statistics book or on the Internet.)

What type of distribution follows the sampling distribution of the means?

A general rule of thumb is that this distribution (the sampling distribution of the means) follows a <em>normal distribution</em> if the sample size, $\\ n$ , is bigger than or equal to 30 observations, or $\\ n \geq 30$ . In this case, $\\ n = 109$ batteries. This is a result from the Central Limit Theorem, fundamental in Statistical Inference.

Standard Deviation

We have to remember that the standard deviation is the square root of the variance $\\ \sigma^2$ , or $\\ \sqrt{\sigma^2}$ .

$\\ \sigma^{2} =5929$
$\\ \sigma = \sqrt{5929} = 77$

Therefore, the standard deviation in this case is $\\ \sigma = 77$ minutes.

In sum, we have the following information to answer this question:

$\\ \sigma = 77$ minutes.
$\\ \mu = 941$ minutes.
$\\ n = 109$ batteries (the sample size is <em>large enough</em> to assume that the sampling distribution of the means follows a <em>normal distribution</em>).
$\\ \overline{x} = 948.8$ minutes.

What is the probability that the mean battery life would be greater than 948.8 minutes?

Well, having all the previous information, we can use [2] to solve this question (without using units):

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{948.8 - 941}{\frac{77}{\sqrt{109}}}$

$\\ z = \frac{7.8}{\frac{77}{\sqrt{109}}}$

$\\ z = \frac{7.8}{7.37526}$

$\\ z = 1.05758 \approx 1.06$

This result is the <em>standardized value</em> or <em>z-score</em> for $\\ \overline{x}$ , considering $\\ \mu = 941$ and $\\ \sigma = 77$ .

We round <em>z</em> to two decimals digits since <em>standard normal table</em> only uses it as an entry to find probabilities.

With $\\ z = 1.06$ , we can consult the <em>cumulative standard normal table. </em>First, we need to find with $\\ z = 1.0$ in the first column in the table. Then, in its first raw, we need to find +0.06. The intersection for these two values determines the cumulative probability for $\\ P(z$ .

It is important to recall that $\\ P(z$ because $\\ z = 1.06$ is the standardized value for $\\ \overline{x} = 948.8$ minutes.

Then, $\\ P(z$

However, the question is about $\\ P(\overline{x} > 948.8) = P(z>1.06)$

And

$\\ P(\overline{x} > 948.8) + P(\overline{x} < 948.8) = 1$

Or

$\\ P(z>1.06) + P(z$

Then

$\\ P(z>1.06) = 1 - P(z$

$\\ P(z>1.06) = 1 - 0.8554$

$\\ P(z>1.06) = 0.1446$

Therefore, "the probability that the mean battery life would be greater than 948.8 minutes" is 0.1446.

6 0

4 years ago

If n represents an odd interge, write expressions in terms of n that represents the next three consecutive odd integers. If the

Wewaii [24]

next three are n+2, n+4, and n+6

four consecutive odd have sum 56

n + n+2 + n+4 + n+6 = 56

4n + 12 = 56

4n = 44

n = 11

the numbers are 11, 13, 15, and 17

6 0

3 years ago