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2 years ago

Is 7.2 bigger than 7.21

Mathematics

2 answers:

Zepler [3.9K]2 years ago

6 0

Answer:

yes 7.2 is bigger than 7.21

Step-by-step explanation:

.2 is 2/10 and .21 is 21/100 which is much smaller that .2

Paraphin [41]2 years ago

4 0

Answer:

yes

Step-by-step explanation:

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Given the following two solids. Complete the following prompts using complete sentences. Write a hypothesis that compares the vo

Bess [88]

Now, there are many types of solids. Hypothesis can be stated as there is a significant discrepancy between solid one from the other sold. Moreover, if the results seems to support the hypothesis then it is somehow reliable and accurate.Thank you for your question. Please don't hesitate to ask in Brainly your queries.

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2 years ago

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"Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad lengt

Westkost [7]

Answer:

a) $\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})$

b) $Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5$

c) $P(\bar X >31.25)=0.006=0.6\%$

d) $P(\bar X >28.25)=0.9997=99.97\%$

e) $P(28.25$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =30,\sigma =2)$

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})$

b. What is the standard error of the mean time?

The standard error is given by this formula:

$Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5$

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(\bar X >31.25)=1-P(\bar X$

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(\bar X >28.25)=1-P(\bar X$

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

$P(28.25$

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2 years ago

A bicycle factory installs about 400 tires per day. Tires are installed Monday through Friday for 8 hours per day. The manager o

lions [1.4K]

Answer:

The results are likely invalid because it is unlikely that the tires installed in one hour on Monday will represent the entire population of tires.

Step-by-step explanation:

Out of 8 hours, do you really think one hour will represent it all?

This is why the resutls are likely invalid ( also I took the quiz )

Answer: The results are likely invalid because it is unlikely that the tires installed in one hour on Monday will represent the entire population of tires.

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2 years ago

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two friends went rowing on the river. they traveled with current for 3 hours. the way back took them 4 hours and 48 minutes. fin

N76 [4]

Answer:

1.5 mph

Step-by-step explanation:

The speed in the still water = 6.5 mph.

The speed of the current = x mph.

1. Two friends traveled with current for 3 hours. The speed with the current is 6.5 + x mph, so they traveled $3(x+6.5)$ miles.

2. Two friends traveled against the current for 4 hours and 48 minutes that is $4\dfrac{4}{5}=4.8$ hours. The speed against the current is 6.5 - x mph, so they traveled $4.8(6.5-x)$ miles.