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earnstyle [38]
4 years ago
13

Two ad agencies want to understand issues that are important to students. Agency A will interview 50 students whose names are ra

ndomly chosen from school enrollment data. Agency B will interview 50 randomly selected people from those who respond to a social media posting. Which method will produce a fair sample of the student population and why?
A. Agency A, because students are convenient and easy to interview
B. Agency A, because names randomly selected from school data will be student names
C. Agency B, because people who respond to social media postings have strong opinions
D. Agency B, because people who respond to social media postings are usually students
Mathematics
1 answer:
cluponka [151]4 years ago
7 0
These ad agencies must focus on their target audience, which are the students. Hence, they should gather data on the pool that will surely comprise of students. For agency B, social media posting is not a good source pool. It's true that students are very participative and opinionated in social media. However, they can't be sure that these are students. Some parents are in social media, as well. Some are working individuals, and some are out of school youth. Unlike agency A, agency B has to sort out profiles first and identify which ones are students. Hence, agency A will produce a fair sample of the student population because it is unarguably true that everyone in the school enrollment data are students.

The answer is B.
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The population of bacteria doubles every 6 hours. By how much will it grow in three days. Write the equation
ch4aika [34]

Answer:

Step-by-step explanation:

let x be the population

so we have ,

2x= 6h

?= 3 days or 3×24h

by cross multiplication we have

2x× 72h= 6h×?

?= 2x×72/6 = 24x

So, in three days the bacteria will b 24 times current population.

8 0
3 years ago
a trucker drove 500 miles in h hours. If the trucker averaged 60 mph, which linear equation could be used to find the number of
irina [24]

Step-by-step explanation:

Given that,

Distance covered by a trucker, d = 500 miles

Average speed of the trucker, v = 600 mph

A trucker drove 500 miles in h hours. We need to write the linear equation used to find the number of hours that the trucker drove. Average speed is given by :

v=\dfrac{d}{t}

t is time in hours

t=\dfrac{d}{v}\\\\t=\dfrac{500\ miles}{60\ mph}

If t is h hours,

h=\dfrac{500\ miles}{60\ mph}\\\\500=60h

On solving, h = 5 hours.

Hence, this is the required solution.

5 0
3 years ago
Given: ΔABC, CM intersects AB<br><br>BC = 5, AB = 7<br><br>CA = 4<img src="https://tex.z-dn.net/?f=%5Csqrt2" id="TexFormula1" ti
finlep [7]

Based on the calculations, the length of side CM is equal to 4 units.

<h3>How to calculate the length of side CM?</h3>

First of all, we would determine the measure of angle B by applying the law of cosine as follows:

B² = A² + C² - 2(A)(C)cosB

<u>Given the following data:</u>

A = BC = 5

B = AC = 4√2

C = AB = 7

Substituting the given parameters into the formula, we have;

(4√(2))² = 5² + 7² - 2 × 5 × 7 × cosB

cos(B) = ((4√(2))² - (5² + 7²))/(-2×5×7)

cos(B) = 0.6

B = cos⁻¹(0.6)

B = 53.13°.

Now, we can find side CM by using sine trigonometry:

sin(B) = CM/BC

CM = BC × sin(B)

CM = 5 × sin(53.13°)

CM = 5 × 0.8

CM = 4 units.

Read more on cosine law here: brainly.com/question/11000638

#SPJ1

4 0
2 years ago
I need help plzzzzzzzzzz
nikklg [1K]
D. (-3,4)

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8 0
3 years ago
PROBLEM SOLVING Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during
Akimi4 [234]

Answer:

Probability that at most 50 seals were observed during a randomly chosen survey is 0.0516.

Step-by-step explanation:

We are given that Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during each  survey.

The numbers of seals observed were normally distributed with a mean of 73 seals and a standard deviation of 14.1 seals.

Let X = <u><em>numbers of seals observed</em></u>

The z score probability distribution for normal distribution is given by;

                             Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean numbers of seals = 73

            \sigma = standard deviation = 14.1

Now, the probability that at most 50 seals were observed during a randomly chosen survey is given by = P(X \leq 50 seals)

            P(X \leq 50) = P( \frac{X-\mu}{\sigma} \leq \frac{50-73}{14.1} ) = P(Z \leq -1.63) = 1 - P(Z < 1.63)    

                                                          = 1 - 0.94845 = <u>0.0516</u>

The above probability is calculated by looking at the value of x = 1.63 in the z table which has an area of 0.94845.

4 0
4 years ago
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