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2 years ago

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The temperature, t, in Burrtown starts at 21°F at midnight, when h = 0. For the next few hours, the temperature drops 4 degrees

every hour.
Which equation represents the temperature, t, at hour h?

A. t = 4h + 21
B. t = –21h + 4
C. t = –4h + 21
D. t = 21h + 4

2 answers:

aleksandr82 [10.1K]2 years ago

8 0

C.
The starting temperature is 21 degrees, so this is the y-intercept.

deff fn [24]2 years ago

7 0

C.
The starting temperature is 21 degrees, so this is the y-intercept.

Since the temperature dropping, the slope of 4 is negative.

The result in slope-intercept form is -4h+21

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Given f(x) = x3 – 2x2 – x + 2, the roots of f(x) are

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2 years ago

Read 2 more answers

−12x−5y=40 12x−11y=88 x= y=

Nezavi [6.7K]

Add the two equations and use the elimination method. The x term will "cancel out" to be able to solve for the y term.

STEP 1
−12x−5y=40
12x−11y=88
x terms "cancels out"; add y terms

-16y= 128
divide both sides by -16

y= -8

STEP 2
substitute y value into either original equation

12x−11y=88

12x - 11(-8)= 88

12x + 88= 88
subtract 88 from both sides

12x= 0
divide both sides by 12; 0 divided by any number is 0

x= 0

ANSWER: x= 0; y= -8

Hope this helps! :)

6 0

2 years ago

Melissa wants to make a table representing the area of her vegetable garden for a variety of different lengths of the tomato pat

Ksenya-84 [330]

Answer:

$\begin{array}{cc}Length \ of \ Tomato \ Patch \ (in \ feet)&Area \ of \ Vegetable \ Garden \ (in \ square \ feet)\\\ [6.25]&338\\6.5&[242.625]\\ \ [6.75]&147.25\\7&[51.875]\end{array}$

Step-by-step explanation:

The date from the given table, is expressed as follows;

Area of Vegetable Garden (in square feet); 338, ___, 147.25, ___

Length of Tomato Patch (in feet); ___, 6.5, ___, 7

The length of the tomato patch increases with decrease in the area of the garden

Given that the length of the tomato patch is the independent variable, we can have;

Length of Tomato Patch (in feet); 6.25, 6.5, 6.75, 7

Therefore, for an increase in the length of the tomato patch from 6.25 to 6.75, (a change of Δl = 0.5) the area of the vegetable garden decreased from 338 to 147.25 which is a decrease of ΔA = 190.75

Therefore, the width of the tomato patch, w = ΔA/Δl = 190.75/0.5 = 381.5

The width of the tomato patch, w = 381.5 ft.

The relationship between the total area, TA, the area of the vegetable garden, <em>A</em>, and the length of the tomato patch, <em>l</em>, is therefore, given as follows;

A = TA - 381.5·l

338 = TA - 381.5×6.25

TA = 338 + 381.5×6.25 = 2,722.375

Therefore, when l = 6.5, we get

A = 2,722.375 - 381.5×6.5 = 242.625

When l = 7, we get

A = 2,722.375 - 381.5×7 = 51.875

Therefore, we get;

$\begin{array}{cc}Length \ of \ Tomato \ Patch \ (in \ feet)&Area \ of \ Vegetable \ Garden \ (in \ square \ feet)\\\ [6.25]&338\\6.5&[242.625]\\ \ [6.75]&147.25\\7&[51.875]\end{array}$

5 0

1 year ago

1. The following are the number of hours that 10 police officers have spent being trained in how to handle encounters with peopl

dusya [7]

Answer:

$Range = 16$

$Inter\ Quartile\ Range = 6.75$

$Variance = 20.44$

$Standard\ Deviation = 4.52$

Step-by-step explanation:

Given

4, 17, 12, 9, 6, 10, 1, 5, 9, 3

Calculating the range;

$Range = Highest - Lowest$

From the given data;

Highest = 17 and Lowest = 1

Hence;

$Range = 17 - 1$

$Range = 16$

Calculating the Inter-quartile Range

Inter quartile range (IQR) is calculates as thus

$IQR = Q_3 - Q_1$

Where

Q3 = Upper Quartile and Q1 = Lower Quartile

<em />

<em>Start by arranging the data in ascending order</em>

1, 3, 4, 5, 6, 9, 9, 10, 12, 17

N = Number of data; N = 10

---------------------------------------------------------------------------------

Calculating Q3

$Q_3 = \frac{3}{4}(N+1) th\ item$

<em>Substitute 10 for N</em>

$Q_3 = \frac{3}{4}(10+1) th\ item$

$Q_3 = \frac{3}{4}(11) th\ item$

$Q_3 = \frac{33}{4} th\ item$

$Q_3 = 8.25 th\ item$

Express 8.25 as 8 + 0.25

$Q_3 = (8 + 0.25) th\ item$

$Q_3 = 8th\ item + 0.25 th\ item$

Express 0.25 as fraction

$Q_3 = 8th\ item +\frac{1}{4} th\ item$

$Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)$

From the arranged data;

$8th\ item = 10$ and $9th\ item = 12$

$Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)$

$Q_3 = 10 +\frac{1}{4} (12 - 10)$

$Q_3 = 10 +\frac{1}{4} (2)$

$Q_3 = 10 +0.5$

$Q_3 = 10.5$

Calculating Q1

$Q_1 = \frac{1}{4}(N+1) th\ item$

<em>Substitute 10 for N</em>

$Q_1 = \frac{1}{4}(10+1) th\ item$

$Q_1 = \frac{1}{4}(11) th\ item$

$Q_1 = \frac{11}{4} th\ item$

$Q_1 = 2.75 th\ item$

Express 2.75 as 2 + 0.75

$Q_1 = (2 + 0.75) th\ item$

$Q_1 = 2nd\ item + 0.75 th\ item$

Express 0.75 as fraction

$Q_1 = 2nd\ item +\frac{3}{4} th\ item$

$Q_1 = 2nd\ item +\frac{3}{4} (3rd\ item - 2nd\ item)$

From the arranged data;

$2nd\ item = 3$ and $3rd\ item = 4$

$Q_1 = 3 +\frac{3}{4} (4 - 3)$

$Q_1 = 3 +\frac{3}{4} (1)$

$Q_1 = 3 +0.75$

$Q_1 = 3 .75$

---------------------------------------------------------------------------------

Recall that

$IQR = Q_3 - Q_1$

$IQR = 10.5 - 3.75$

$IQR = 6.75$

Calculating Variance

Start by calculating the mean

$Mean = \frac{1+3+4+5+6+9+9+10+12+17}{10}$

$Mean = \frac{76}{10}$

$Mean = 7.6$

Subtract the mean from each data, then square the result

$(1 - 7.6)^2 = (-6.6)^2 = 43.56$

$(3 - 7.6)^2 = (-4.6)^2 = 21.16$

$(4 - 7.6)^2 = (-3.6)^2 = 12.96$

$(5 - 7.6)^2 = (-2.6)^2 = 6.76$

$(6 - 7.6)^2 = (-1.6)^2 = 2.56$

$(9 - 7.6)^2 = (1.4)^2 = 1.96$

$(9 - 7.6)^2 = (1.4)^2 = 1.96$

$(10 - 7.6)^2 = (2.4)^2 = 5.76$

$(12 - 7.6)^2 = (4.4)^2 = 19.36$

$(17 - 7.6)^2 = (9.4)^2 = 88.36$

Sum the result

$43.56 + 21.16 + 12.96 + 6.76 + 2.56 + 1.96 + 1.96 + 5.76 + 19.36 + 88.36 = 204.4$

Divide by number of observation;

$Variance = \frac{204.4}{10}$

$Variance = 20.44$

Calculating Standard Deviation (SD)

$SD = \sqrt{Variance}$

$SD = \sqrt{20.44}$

$SD = 4.52$ <em>(Approximated)</em>

4 0

2 years ago