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Dimas [21]
2 years ago
7

The scatterplot shows a linear relationship between the distance traveled and the time elapsed. What is the rate of change of th

e linear relationship?
For every increase in time by 1 hour, the distance increases
36 miles.
54 miles.
72 miles.
90 miles.

Mathematics
2 answers:
emmainna [20.7K]2 years ago
7 0
To solve the question we get the slope for the given  graph. This is given by:
slope,m=(change in y-values)/(change in x-values)
taking 2 points (1.25, 90) and (0.75,54) we get:
slope,m=(90-54)/(1.25-0.75)=36/0.5=72
The answer is:
72 miles
lukranit [14]2 years ago
5 0

Answer:

The answer is C. 72 miles.

Step-by-step explanation:

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Orlov [11]
The whole second is the first number to the left of the decimal (_.), so : 35.78=36 because 7>5. 39.51=40 because 5=5. 40-36=4. Bailey won the race by about 4 seconds. Hoped I helped!
8 0
2 years ago
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susan took two tests.the probability of her passing both tests is 0.6.the probability of her passing the first test is 0.8.what
OverLord2011 [107]
Formula for this is as follows:
probability of her passing both 0.6/0.8 - first test and this is a fraction. 0.6/0.8
0.6/0.8= divide 0.6 by 0.8=0.75
that means probability of her passing the second test is 75%
6 0
2 years ago
Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ
SOVA2 [1]

y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}

The arc length of the curve is

\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx

which has a value of about 5.99086.

Let f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}. Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

\ell_i=\dfrac{i-1}2

r_i=\dfrac i2

with 1\le i\le10.

These subintervals have midpoints given by

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

Over each subinterval, we approximate f(x) with the quadratic polynomial

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

so that the integral we want to find can be estimated as

\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that

\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6

so that the arc length is approximately

\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086

5 0
2 years ago
Theresa has hired Chuck and Diana to paint a fence. Diana can paint 150 fence posts in the same time it takes Chuck to paint 130
Alenkasestr [34]

Answer: Chuck can paint 65 posts per hour.

Step-by-step explanation:

The data we have here is:

Diana can paint 150 fence posts in a time T

Chuck can pint 130 fence posts in a time T.

In one hour, Diana can paint 10 more fence posts than Chuck.

If D is the hourly rate of Daian, and C is the hourly rate for Chuck, we have:

D = 150/T

C = 130/T

D = C + 10

we can replace the last equation in the first one and geT:

C + 10 =150/T

C = 130/T

Now we can replace the bottom equation in the above one:

130/T + 10 = 150/T

10 = 150/T - 130/T =  20/T

T*10 = 20

T = 20/10 = 2

So T is 2 hours, then whe have:

D = 150post/2hours = 75 posts per hour.

C = 130 posts/2hours = 65 posts per hour.

4 0
2 years ago
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atroni [7]
-7 and 8 are the solutions to the given equation system.
Therefore, the maximum distance between the y values of the two equations must lie exactly between their points of intersection. That is on x value:
x = (-7 + 8)/2 = 0.5
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y = 0.5 + 56 = 56.5
y = 0.5² = 0.25
56.5 - 0.25 = 56.25 units
5 0
2 years ago
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