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lozanna [386]
3 years ago
15

A survey of 120 twelfth-graders finds that 36% were carrying more than $15 on the day of the survey. Use the margin of error to

complete the following. Round to the nearest whole percent. An interval that will likely include the proportion of students in the population of twelfth-graders who carry more than $15 is ?
Mathematics
1 answer:
-Dominant- [34]3 years ago
3 0

Answer:

An interval that will likely include the proportion of students in the population of twelfth-graders who carry more than $15 is 960

Step-by-step explanation:

For example, Condition 1: n(.05)≤N

• The sample size (10) is less than 5% of the population (millions of musicians), so

the condition is met.

• Condition 2: np(1-p)≥10

• =

2

10

= .2

• 1 − = 10 .2 1 − .2 = 1.6 . This is less than 10 so this condition is not

met.

It would not be practical to construct the confidence interval.

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wlad13 [49]

The first and third option.

Hope this helps.

7 0
3 years ago
PLEASE help me ASAP i give brainliest!
faltersainse [42]

Answer:

scatter plot A: -0.90

scatter plot B: 0.89

scatter plot C: -0.76

scatter plot D: 0.55

Step-by-step explanation:

if the line is going down, it's negative and if it goes up it's positive. The closer the points are <em>the higher the number</em>.

HOPE THIS HELPS!!

3 0
3 years ago
Someone please help me
Lilit [14]
<h3>Answer:</h3>
  1. C. (9x -1)(x +4) = 9x² +35x -4
  2. B. 480
  3. A. P(t) = 4(1.019)^t

Step-by-step explanation:

1. See the attachment for the filled-in diagram. Adding the contents of the figure gives the sum at the bottom, matching selection C.

2. If we let "d" represent the length of the second volyage, then the total length of the two voyages is ...

... (d+43) + d = 1003

... 2d = 960 . . . . . . . subtract 43

... d = 480 . . . . . . . . divide by 2

The second voyage lasted 480 days.

3. 1.9% - 1.9/100 = 0.019. Adding this fraction to the original means the original is multiplied by 1 +0.019 = 1.019. Doing this multiplication each year for t years means the multiplier is (1.019)^t.

Since the starting value (in 1975) is 4 (billion), the population t years after that is ...

... P(t) = 4(1.019)^t

7 0
3 years ago
prove that if f is a continuous and positive function on [0,1], there exists δ &gt; 0 such that f(x) &gt; δ for any x E [0,1] g
ValentinkaMS [17]

Answer:

I dont Know

Step-by-step explanation:

5 0
3 years ago
Y=3/2(-1)-1<br> Linear Equation
levacccp [35]
X = -5/2
X is a single horizontal line across the graph, because x has a set value

5 0
4 years ago
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