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lord [1]
3 years ago
12

• A research scientist measures the outside temperature, in degrees Fahrenheit, at various

Mathematics
1 answer:
polet [3.4K]3 years ago
5 0

Answer:

{t|60 <= t <= 85}

Step-by-step explanation:

The temperatures were measures at different times, but does not stop the values being real numbers (i.e. not discrete, or integer values).

So the range of the function is the set of all values between the minimum and maximum measured during the measuring interval (domain) of hours two and twenty-two.

The minimum value = 60F

The maximum value = 85F

So the interval of the range is [60,85], in interval notation.

In set-builder notation, it is

{t|60 <= t <= 85}

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Which method correctly solves the equation using the multiplication property of equality and the reciprocal of One-third? One-th
Airida [17]

Answer:

Step-by-step explanation:

Given the equation \frac{1}{3}(x+9) = -12

Step 1;

Expand the bracket at the right hand side of the equation to have:

\frac{1}{3}x +\frac{1}{3}(9) = -12\\\frac{1}{3}x+3=-12\\subtracting\ 3\ from\ both\ sides\\\frac{1}{3}x+3-3=-12-3\\\frac{1}{3}x=-15\\

Taking the reciprocal of both sides:

\frac{3}{x} = \frac{-1}{15} \\ cross\ multiplying\\-x =45\\x=-45

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3 years ago
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The following information is given for Tripp Company, which uses the indirect<br> method
oee [108]

Answer:

Step-by-step explanation:

1.

The cash flow from operating activities is = $28000.

Working Note:

Net Income                                                        $20000

Depreciation expense                                   $3000

Increase in accounts receivable                 ($2000)

Increase in accounts payable                      $4000

Decrease in inventory                                    $3000

Net Cash Flow from Operating Activities = 20000 + 3000 – 2000 + 4000 + 3000

Net Cash Flow from Operating Activities = $28000

2.

The cash flow from investing activities = $6000.

Working Note:

Proceeds from sale of equipment                            $6000

Net cash flow from investing activities =               $6000

3.

The cash flow from financing activities = - $2000              or ($2000)

Working Note:

Payment of dividends                                    ($2000)

Net Cash flow from financing activities = - $2000                or ($2000)

7 0
3 years ago
There were 29 students available for the woodwind section of the school orchestra. 11 students could play the flute, 15 could pl
Dafna11 [192]

Answer:

a. The number of students who can play all three instruments = 2 students

b. The number of students who can play only the saxophone is 0

c. The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only one of the clarinet, saxophone, or flute = 4

Step-by-step explanation:

The total number of students available = 29

The number of students that can play flute = 11 students

The number of students that can play clarinet = 15 students

The number of students that can play saxophone = 12 students

The number of students that can play flute and saxophone = 4 students

The number of students that can play flute and clarinet = 4 students

The number of students that can play clarinet and saxophone = 6 students

Let the number of students who could play flute = n(F) = 11

The number of students who could play clarinet = n(C) = 15

The number of students who could play saxophone = n(S) = 12

We have;

a. Total = n(F) + n(C) + n(S) - n(F∩C) - n(F∩S) - n(C∩S) + n(F∩C∩S) + n(non)

Therefore, we have;

29 = 11 + 15 + 12 - 4 - 4 - 6 + n(F∩C∩S) + 3

29 = 24 + n(F∩C∩S) + 3

n(F∩C∩S) = 29 - (24 + 3) = 2

The number of students who can play all = 2

b. The number of students who can play only the saxophone = n(S) - n(F∩S) - n(C∩S) - n(F∩C∩S)

The number of students who can play only the saxophone = 12 - 4 - 6 - 2 = 0

The number of students who can play only the saxophone = 0

c. The number of students who can play the saxophone and the clarinet but not the flute = n(C∩S) - n(F∩C∩S) = 6 - 2 = 4

The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only the saxophone = 0

The number of students who can play only the clarinet = n(C) - n(F∩C) - n(C∩S) - n(F∩C∩S) = 15 - 4 - 6 - 2 = 3

The number of students who can play only the clarinet = 3

The number of students who can play only the flute = n(F) - n(F∩C) - n(F∩S) - n(F∩C∩S) = 11 - 4 - 4 - 2 = 1

The number of students who can play only the flute = 1

Therefore, the number of students who can play only one of the clarinet, saxophone, or flute = 1 + 3 + 0 = 4

The number of students who can play only one of the clarinet, saxophone, or flute = 4.

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The answer to this question is $57.02
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3 years ago
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alekssr [168]
Part A:

3.14 X (4)^2 X 3/3
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