Which is the longest? A. 2 km B. 25 m C. 2,500 cm D. 3,000 mm

The cost for fudge at Candy Lane is represented by the table. The graph shows the cost of a similar fudge that Best Fudge Shop s

Sedbober [7]

Answer:

B

Step-by-step explanation:

See attached image for table and graph.

Looking at Candy Lane, we can see that

1 cost 8

2 cost 16

3 cost 24

....etc

That tells us that unit rate is $8

Now, looking at graph of Best Fudge, we can see that

at x = 1, y = 3

at x = 2, y = 6

...

etc.

So we can say unit rate is $3.

Certainly, unit rate of Best Fudge is less than unit rate of Candy Lane. Answer choice B is right.

6 0

4 years ago

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The lengths of a particular snake are approximately normally distributed with a given mean mc025-1.jpg = 15 in. and standard dev

Vedmedyk [2.9K]

Answer:

2.5% of the snakes are longer than 16.6 in

Step-by-step explanation:

The Z score is,

$Z=\dfrac{X-\mu}{\sigma}$

here,

X = raw score = 16.6

μ = mean = 15

σ = standard deviation = 0.8

So,

$Z=\dfrac{16.6-15}{0.8}=2$

Z-score is the measurement of how many standard deviation away the data is from the mean. Hence, the value 16.6 is 2 standard deviations right to the mean.

Approximately 95% of the data falls within two standard deviations of the mean. (between the mean-2 times the standard deviation, and the mean+2 times the standard deviation)

So the 5% of data are left in the remaining space. As we have to find the amount of snakes which are longer than 16.6 in i.e we have to find the area to the right of 16.6.

The remaining 5% of amount is divided in two equal parts.

Therefore, $\dfrac{5}{2}=2.5\%$ of the snakes are longer than 16.6 in

6 0

3 years ago

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Differential Equations Problem

fgiga [73]

Answer:

$y=\frac{t^2e^{2t}}{3}+ce^{2t}$

Step-by-step explanation:

We have given differential equation $\frac{dy}{dt}-2y=t^2e^{2t}$

We know that linear differential equation is given by $\frac{dy}{dt}+Py=Q$

On comparing with standard equation P = -2 and Q= $t^2e^{2t}$

Now integrating factor $IF=e^{-Pdt}$

$IF=e^{-2dt}=e^{-2t}$

Now solution of differential equation is given by

$y\times IF=\int\ IF\times Q\ dt$

$y\times e^{-2t}=\int\ e^{-2t}\times t^2e^{2t}\ dt$

$y\times e^{-2t}=\frac{t^2}{3}+c$

$y=\frac{t^2e^{2t}}{3}+ce^{2t}$

7 0

4 years ago

Complete the equation of the line through (3,-1)(3,−1)left parenthesis, 3, comma, minus, 1, right parenthesis and (4,7)(4,7)left

Mandarinka [93]

Answer: $y=8x-25$

Step-by-step explanation:

Given

Line passes through $(3,-1)\ \text{ and}\ (4,7)$

Equation of line when it passes through two points $(x_1,y_1),(x_2,y_2)$

$\Rightarrow \dfrac{y_2-y_1}{x_2-x_1}=\dfrac{y-y_1}{x-x_1}$

Insert the values

$\Rightarrow \dfrac{7+1}{4-3}=\dfrac{y+1}{x-3}\\\\\Rightarrow \dfrac{8}{1}=\dfrac{y+1}{x-3}\\\\\Rightarrow 8x-24=y+1\\\Rightarrow y=8x-25$

8 0

3 years ago

Determine the equation of the line that is perpendicular to the lines r(t)=(-2+3t,2t,3t)

Mnenie [13.5K]

Vector Equation
(Line)(x,y) = (x,y) + t(a,b);tERParametric Formx = x + t(a), y = y + t(b); tERr = (-4,-2) + t((-3,5);tERFind the vector equation of the line passing through A(-4,-2) & parallel to m = (-3,5)Point: (2,5)
Create a direction vector: AB = (-1 - 2, 4 - 5)
= (-3,-1) or (3,1)when -1 (or any scalar multiple) is divided out.
r = (2,5) + t(-3,-1);tERFind the vector equation of the line passing through A(2,5) & B(-1,4)x = 4 - 3t
y = -2 + 5t
;tERWrite the parametric equations of the line passing through the line passing through the point A(4,-2) & with a direction vector of m =(-3,5)Create Vector Equation first:
AB = (2,8)
Point: (4,-3)
r = (4,-3) + (2,8); tER
x = 4 + 2t
y = -3 + 8t
;tERWrite the parametric equations of the line through A(4,-3) & B(6,5)Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in -3
-3 = 5 + 4t
(-8 - 5)/4 = t
-2 = t
For y sub in -8
-8 = -2 + 3t
(-8 + 2)/3 = t
-2 = t
Parameter 't' is consistent so pt(-3,-8) is on the line.Given the equation r = (5,-2) + t(4,3);tER, is (-3,-8) on the line?Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in 1
-1 = 5 + 4t
(-1 - 5)/4 = t
-1 = t
For y sub in -7
-7 = -2 + 3t
(-7 + 2)/3 = t
-5/3 = t
Parameter 't' is inconsistent so pt(1,-7) is not on the line.Given the equation r = (5,-2) + t(4,3);tER, is (1,-7) on the line?Use parametric equations when generating points:
x = 5 + 4t
y = -2 + 3t ;tER
X-int:
sub in y = 0
0 = -2 + 3t
solve for t
2/3 = t (this is the parameter that will generate the x-int)
Sub t = 2/3 into x = 5 + 4t
x = 5 + 4(2/3)
x = 5 + (8/3)
x = 15/3 + (8/3)
x = 23/3
The x-int is (23/3, 0)What is the x-int of the line r = (5,-2) + t(4,3); tER?Note: if they define the same line: 1) Are their direction vectors scalar multiples? 2) Check the point of one equation in the other equation (LS = RS if point is subbed in)What are the two requirements for 2 lines to define the same line?

3 0

4 years ago