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snow_tiger [21]
1 year ago
7

A researcher says to the respondents in a poll, “Eating too many sugary foods leads to cavities. Would you rather have soda or w

ater served with your meal?” Is this a valid question to ask of sample respondents?
A. Yes, the more information provided by a researcher the better. Respondents can now give an informed opinion and the results will be more accurate.

B. No, the wording of the question makes respondents more likely to say water, even if they may actually prefer soda at a meal.

C. No, a researcher cannot ask people for preferences because they may not answer honestly. The researcher should observe people and record their beverage selections to insure accurate responses.

D. Yes, the researcher is simply stating a fact: eating sugary foods does lead to cavities. It is okay for a researcher to state facts in asking questions of respondents.
Mathematics
1 answer:
mafiozo [28]1 year ago
3 0

The correct answer is B. No, the wording of the question makes respondents more likely to say water, even if they may actually prefer soda at a meal.

Explanation:

One important factor when designing questions in research is to avoid any language that might influence the answers of respondents. This recommendation was not followed in the question "Eating too many sugary foods leads to cavities. Would you rather have soda or water served with your meal?" because mentioning sugary foods, which includes soda, leads to cavities will make respondents consider soda is negative and they are more likely to choose water. This implies the wording in the question influences respondents and introduces bias, which is inappropriate. Thus, the correct answer is B.

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The brass strip AB has been attached to a fixed support at A and rests on a rough support at B. Knowing that the coefficient of
Andre45 [30]

Answer:

Decrease in temperature = 4.6714°C

Step-by-step explanation:

We know that the formula fro deflection is:

δ = PL/AE + LαΔT

As deflection is equal to zero in this case, the formula becomes

ΔT = -P/αAE   -------------- equation (1)

Now,

we see that only vertical force acting is weight

W = mg = 100*9.81

W = 981N = ∑F_{y}                                  where 1N = 1kgm/s²

Now,

for horizontal forces

As we know that, Sum of all horizontal forces = 0

P - μN = 0

P - μW = 0

P = μW

P = 0.6 * 981

P = 588.8N or 589N

Now we need to calculate the area

area = width * thickness = 20 * 3

area = 60 mm²

Now by substituting all these values in equation 1, wee get

ΔT = -P/αAE = - 588.6/(20*10⁻⁶*60*10⁻⁶*105*10⁹)

ΔT = - 4.6714°C (negative sign indicates decrease)

7 0
1 year ago
Ria can paint a room in 4 hours. Destiny can paint the same room in 6 hours. How long would it take Ria and Destiny to paint the
Lerok [7]

Time taken by Ria to paint a room = 4 hours

Time taken by Destiny to paint a room = 6 hours

If they work together, they complete 1 job. So,

\frac{1x}{4}+\frac{1x}{6}=1

The LCD of 4 and 6 is 12

\frac{3x+2x}{12}=1

\frac{5x}{12}=1

5x=12

x=\frac{12}{5}

Hence, both of them can paint the room in \frac{12}{5} hours or 2.4 hours.


4 0
2 years ago
Read 2 more answers
Sam gave jen 1/2 of his jujubes .jen ate 1/2 ofthe jujubes and gave rest to kyle.kyle kept 8 of the jujubes and gave the last 10
erastovalidia [21]
Let x = the # jujubes Sam had
he gave x/2 to jen
jen eats half of it  =    1/2(x/2) =    x/4
the remaining  x/4 he gave to kyle
kyle kept 8 and gave 10 to Kim
so
             x/4 - 8 = 10
             x/4 = 18
             x =72
thus    jen had 36 jujubes
jen eat    18  jujubes
3 0
1 year ago
Consider the midterm and final for a statistics class. Suppose 13% of students earned an A on the midterm. Of those students who
padilas [110]

Answer:

There is a 38.97% probability that this student earned an A on the midterm.

Step-by-step explanation:

The first step is that we have to find the percentage of students who got an A on the final exam.

Suppose 13% students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final.

This means that

Of the 13% of students who earned an A on the midterm, 47% received an A on the final. Also, of the 87% who did not earn an A on the midterm, 11% received an A on the final.

So, the percentage of students who got an A on the final exam is

P_{A} = 0.13(0.47) + 0.87(0.11) = 0.1568

To find the probability that this student earned an A on the final test also earned on the midterm, we divide the percentage of students who got an A on both tests by the percentage of students who got an A on the final test.

The percentage of students who got an A on both tests is:

P_{AA} = 0.13(0.47) = 0.0611

The probability that the student also earned an A on the midterm is

P = \frac{P_{AA}}{P_{A}} = \frac{0.0611}{0.1568} = 0.3897

There is a 38.97% probability that this student earned an A on the midterm.

5 0
1 year ago
In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that
attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable X\sim Poisson(\lambda = 0.62).

The probability function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

             = 1 - P (X = 0) - P (X = 1)

             =1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: \lambda=7\times0.62=4.34

Compute the probability of no deaths in a corps over 7 years as follows:

P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0
1 year ago
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