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Svet_ta [14]
3 years ago
10

In the regression equation y = -1.2x + 102.45, x represents the number of cloudy school days and y represents the percent of sch

ool attendance on those days. Predict the per cent of school attendance during a month with 18 cloudy days. A) 78.52% B) 80.85% C) 90.36% D) 93.47%
Mathematics
1 answer:
MAVERICK [17]3 years ago
4 0

Answer:

B.80.85%

Step-by-step explanation:

The equation shows that the percent of school attendance depends on cloudy days, in a negative way, which means that the more cloudy days, the less the percentage of students that go to school.

  • Particular, when there are 18 cloudy days, x takes a value of 18, x=18,
  • This means that y= -1.2 (18) + 102.45 which equals 80.85%.

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Which two of the following is continuous data?
Alla [95]

Answer:

B. Age of student

D. Time taken to run 1 mile

Step-by-step explanation:

From the list of given options, only B and D satisfy the required condition.

One unique determinant of continuous data is that; they are measured and not counted.

Now, let's categorize option A to D into two

1. Counted data

2. Measured data

Options that fall into the category of measured data are said to be continuous data.

A. Concert attendance; The number of people in a concert is counted

B. The age of a student is measured (in years)

C. Number of pens in a box is counted

D. Time taken to run 1 mile is measured (in units like seconds, minutes, hours, etc...)

In summary; we have

Counted

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C. Number of pens in a box

Measured

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7 0
3 years ago
(08.03)
Alik [6]

Answer:

Option D, (x + 2)(x - 6)

Step-by-step explanation:

<u>Step 1:  Factor </u>

x^2 - 4x - 12

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(x - 6)(x + 2)

(x + 2)(x - 6)

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8 0
3 years ago
How many times longer is the wavelength of a sound wave with a frequency of 20 waves per second than the wavelength of a sound w
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The sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec

<h3>Calculating wavelength </h3>

From the question, we are to determine how many times longer is the first sound wave compared to the second sound water

Using the formula,

v = fλ

∴ λ = v/f

Where v is the velocity

f is the frequency

and λ is the wavelength

For the first wave

f = 20 waves/sec

Then,

λ₁ = v/20

For the second wave

f = 16,000 waves/sec

λ₂ = v/16000

Then,

The factor by which the first sound wave is longer than the second sound wave is

λ₁/ λ₂ = (v/20) ÷( v/16000)

= (v/20) × 16000/v)

= 16000/20

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Hence, the sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec

Learn more on Calculating wavelength here: brainly.com/question/16396485

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6 0
2 years ago
John has 15 t shirts. 6 black, 3 red, 2 purple, 2 dark blue, 1 light blue, 1 purple
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Answer:

ok.... what is the question here....??   ;-;

Step-by-step explanation:

3 0
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NNADVOKAT [17]

Answer:

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4 years ago
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