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Alla [95]
3 years ago
5

A researcher would like to evaluate the claim that large doses of vitamin C can help prevent the common cold. One group of parti

cipants is given a large does of the vitamin (500 mg per day), and a second group is given a placebo (sugar pill). The researcher records the number of colds each individual experiences during the 3-month winter season. Is the dependent variable (number of colds within a winter season) discrete or continuous?
Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
7 0

Answer: The dependent variable is discrete.

Step-by-step explanation:

There are different types of variables. Variables can be classified into two main groups: qualitative and quantitative.

Qualitative variables are those who represent attributes. For example, flowers of 3 different colors: yellow, orange and blue. Also, when they haven’t an order or a hierarchy, they are called nominal, as in the previous flowers's example. When they have a hierarchy associated, they are called ordinal (you can distinguish a natural order), for example: small, medium, large.  

Quantitative variables are those that represent quantities, and they can be discrete or continuous. A variable is discrete when only certain finite values ​​are possible. This variables are used to represent counts, like the in the example: number of colds within a winter season, or for example, number of students in a class, number of plants per acre, etc.

On the other hand, when a variable is continuous, infinite values ​​can be found between two values. They are measure in continuous scales. For example, the plant’s height: it could be take any continuous values, like 10.3 cm, 15 cm, 17.99 cm.

In summary, in the present problem the response variable is quantitative and discrete, because it is countable and finite, and only some values ​​are possible: positive integers.

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Find the lateral area of this pyramid whose base is a regular hexagon with a side length of 3cm and whose slant height is 12cm.
juin [17]

Answer:

108 cm^2

Step-by-step explanation:

The lateral surface area of a pyramid is given as

LSA= 1/2×perimeter of the base×slant height

base is hexagon with side 3 cm , therefore, perimeter of the base

=6×3=18 cm

slant height = 12 cm

thus, LSA = 1/2×18×12 = 108 cm^2

4 0
3 years ago
At a soccer tournament 12 teams are wearing red shirts, 6 teams are wearing blue shirts, 4 teams are wearing orange shirts, and
Alchen [17]
When you add 12, 6, 4 and 2, you will get 24. 12 of the 24 teams wear red shirts, so for every 2 teams there would be 1 team wearing red
8 0
3 years ago
Read 2 more answers
The bad debt ratio for a financial institution is defined to be the dollar values of loans defaulted divided by the total dollar
Nimfa-mama [501]

Answer:

(a) NULL HYPOTHESIS, H_0 : \mu \leq  3.5%

    ALTERNATE HYPOTHESIS, H_1 : \mu > 3.5%

(b) We conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Step-by-step explanation:

We are given that a random sample of seven Ohio banks is selected.The bad debt ratios for these banks are 7, 4, 6, 7, 5, 4, and 9%.The mean bad debt ratio for all federally insured banks is 3.5%.

We have to test the claim of Federal banking officials that the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

(a) Let, NULL HYPOTHESIS, H_0 : \mu \leq  3.5% {means that the the mean bad debt ratio for Ohio banks is less than or equal to the mean for all federally insured banks}

ALTERNATE HYPOTHESIS, H_1 : \mu > 3.5% {means that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks}

The test statistics that will be used here is One-sample t-test;

                T.S. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where,  \bar X = sample mean debt ratio of Ohio banks = 6%

             s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} } = 1.83%

             n = sample of banks = 7

So, test statistics = \frac{6-3.5}{\frac{1.83}{\sqrt{7} } }  ~ t_6

                             = 3.614

(b) Now, at 1% significance level t table gives critical value of 3.143. Since our test statistics is more than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Hence, Federal banking officials claim was correct.

7 0
3 years ago
Edmentum// monomials
stepladder [879]

Answer:

625b^6

Step-by-step explanation:

= 5^3 * 5b^6

= 5^3 * 5

= 625

6 0
3 years ago
Evaluate 5+(-3) — 6.
WARRIOR [948]

Answer:

-4

Step-by-step explanation:

5+(-3)-6

2-6

-4

4 0
3 years ago
Read 2 more answers
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