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3 years ago

7

A water rocket is projected vertically upward from the ground. Its a height(h) m above the ground after t seconds is given by h=

20-5(t-2)^2. When will the rocket reach the ground?
MARKING THE BRAINLIEST

1 answer:

Morgarella [4.7K]3 years ago

5 0

The rocket will reach the ground after 4 seconds

Step-by-step explanation:

Step 1: Find the time when the water rocket reaches the ground.

When the water rocket reaches the ground, the height, h will be zero.

⇒ h = 20 - 5(t - 2)²

⇒ 0 = 20 - 5(t - 2)²

⇒ 5 (t - 2)² = 20

⇒ (t - 2)² = 4

⇒ t - 2 = √4 = 2

∴ t = 2 + 2 = 4 seconds

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Step-by-step explanation:

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"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:

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

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

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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):

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

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

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

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

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

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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

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

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

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

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