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Phoenix [80]
2 years ago
8

A survey of 898 U.S. VCR owners found that 16% had VCR clocks that were currently blinking “12:00.” (Source: Wirthlin Worldwide)

The population is the 898 owners. true or false
Mathematics
2 answers:
yKpoI14uk [10]2 years ago
8 0

In this case scenario, the population and sample would be:

Population : Collection of all US adult who own VCR

Sample : Collection 898 US adult who own VCR

<span>Therefore the answer to this is “False”</span>

<span>Hope this helps you!</span>

musickatia [10]2 years ago
8 0

Answer:

This is false. 898 owners are sample.

Step-by-step explanation:

Given condition is -

A survey of 898 U.S. VCR owners found that 16% had VCR clocks that were currently blinking “12:00.”

Now, population is defined as the collection of all outcomes, responses and counts that are of interest in the survey.

Here, we can define the following:

Population: U.S. adult VCR owners

Sample: 898 owners

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What is the difference between 5 hundreds and 3 tens?​
Umnica [9.8K]

Answer:

5 tens and 3 hundreds = 50 + 300 = 350. That is basically asking what number goes 5-3-2. This is because 5 hundreds means 5 in the hundreds place, or 500. 3 tens is just saying 3 in the tens place, or 30.

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Can anyone help me understand how to evaluate the limit of this complex fraction?
SSSSS [86.1K]

Answer:

-1/9

Step-by-step explanation:

\lim_{x \to 3} \frac{1/x-1/3}{x-3}

For simplicity, let's multiply top and bottom by 3x:

\lim_{x \to 3} \frac{3-x}{3x(x-3)}

Factor out a -1:

\lim_{x \to 3} \frac{-(x-3)}{3x(x-3)}

Divide top and bottom by x−3:

\lim_{x \to 3} \frac{-1}{3x}

Evaluate the limit:

\frac{-1}{3(3)}\\-\frac{1}{9}

It's important to note that the function doesn't exist at x = 3.  As x <em>approaches</em> 3, the function <em>approaches</em> -1/9.

5 0
3 years ago
301 345 305 394 328 317 361 325 310 362 What values of minimum, Q1, median, Q3, and maximum should be used to make a box plot fo
Hunter-Best [27]
First, put the numbers in order...very important
301,305,310,317,325,328,345,361,362,394

minimum (smallest number) = 301
Q1 = 310
Q2 (middle number) = (325 + 328)/2 = 653/2 = 326.5
Q3 = 361
maximum (largest number) = 394

3 0
3 years ago
For which equation would x=5 be a solution?
kari74 [83]
03x+6=21

3(5) = 15

15+6=21
8 0
3 years ago
Find the absolute maximum and absolute minimum values of f on the given interval. (If an answer does not exist, enter DNE.)
vodka [1.7K]

Answer:

absolute max is 120 and absolute min is -8

Step-by-step explanation:

Find critical numbers

f'(x) = 3x^2 - 12x + 9 = 0

= 3(x^2 - 4x + 3) = 0

3(x-3)(x-1) = 0

(x-3) = 0 or (x-1)=0

x = 1,3

Test them!

x<1     Sign of f' on this interval is positive

1<x<3 Sign of f' on this interval is negative

x>3    Sign of f' on this interval is positive

f(x) changes from positive to negative at x = 1 which means there is a relative maximum here.

f(x) changes from negative to positive at x = 3 which means there is a relative minimum here.

Test the endpoints to find the absolute max and min.

f(-1) = -8

f(1) = 12

f(3) = 8

f(7) = 120

The absolute maximum value of f is 120 and the absolute minimum value of f is -8.

7 0
3 years ago
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