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3 years ago

SOMEONE PLEASE HELP ME!!!!! I NEED HELP ON CASE #3

Mathematics

1 answer:

sweet-ann [11.9K]3 years ago

4 0

I feel like I've answered this already, but I could be wrong. Since it took 10 years for the price to increase by 11 cents, it would take around 7-8 years for it to raise by .8. (.52- .44)

Hope this helps!

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Brent counted 5 red cards, 10 black cards, and 20 blue cards in a deck of cards. What is the ratio of blue cards to red cards?

Elena-2011 [213]

Answer:

20:5 - If you're going to simplify it, do 20/5 which is 4/1.

Step-by-step explanation:

6 0

2 years ago

1. A study of college freshmen found that students who watched TV for more than one hour each night gained 15 pounds in the firs

umka21 [38]

Answer:

C. They most likely used an observational study asking students to keep track of their TV habits and weight.

A. This is an experiment because a treatment was applied to a group.

Step-by-step explanation:

According to our question:

1. The study concluded that there is a relation between watching TV and the weight gain in college freshmen.

Now, this is possible when a group of college students were asked to participate to watch TV for more than 1 hour for a specific amount of time and have a record of their weights.

So, the study is an observational study where the habits and weights of college students were observed.

Hence, option C is correct.

2. The students were divided onto two groups where one group used the computer program and the other did not to check the improvement in the reading levels.

This means that the students were given a procedure to follow during the study in order to see the results..

So, the study is an experiment where a treatment is applied to the group of students.

Hence, option A is correct.

8 0

2 years ago

Read 2 more answers

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$