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borishaifa [10]
2 years ago
13

What is the slope and y-intercept of the equation represented by the following graph

Mathematics
1 answer:
Mandarinka [93]2 years ago
6 0

Answer: C slope= 1/2 y-intercept =2

Step-by-step explanation:

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Suppose flights on standard routes between two given cities use on average 3,087 gallons of kerosene with standard deviation of
Radda [10]

Answer:

15.866%

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 3,282 gallons of kerosene

μ is the population mean = 3,087 gallons of kerosene

σ is the population standard deviation = 195 gallons

For x > 3282 gallons

z = 3282 - 3087/195

z = 1

Probability value from Z-Table:

P(x<3282) = 0.84134

P(x>3282) = 1 - P(x<3282) = 0.15866

Converting to percentage

0.15866 × 100 = 15.866%

The percent of flights on these particular routes that will burn more than 3,282 gallons of kerosene is 15.866%

3 0
2 years ago
The function f(x) is defined below. What is the end behavior of f(x) ?
Dominik [7]
1363.xs and nun of yourrrt bennies
8 0
3 years ago
WILL GIVE BRAINLIEST TO CORRECT ANSWER
Musya8 [376]
The volume of a Cone = \frac{1}{3} \pi r^{2}h

Since for the ice cream cone has h = 4 inches; r = 0.75; π = 3.14

Then Volume of the Cone = 2.355 in³

∴   This ice cream cone can hold  ≈  2.36 in³ of ice cream without overflowing
7 0
3 years ago
Read 2 more answers
What is the equation of a line with a slope of 4 and an x-intercept of -10
VladimirAG [237]

Answer:

y  = 4x + 40

Step-by-step explanation:

y - 0 = 4 (x + 10)

y -0 = 4x + 40

y = 4x + 40

5 0
3 years ago
Matlab the equation of a circle with its center at x=3 and y=2 is given by , where r is the radius of the circle. first derive t
Blizzard [7]
The equation of a circle with center at (h,k) and radius r is
(x-h)^2+(y-k)^2=r^2
we are given
center is at (3,2)
(x-3)^2+(y-2)^2=r^2
solving for y
(y-2)^2=-1(x-3)^2+r^2
y-2=\sqrt{-1(x-3)^2+r^2}
y=2+\sqrt{-1(x-3)^2+r^2}
take the derivitive to find the slope at any point
\frac{dy}{dx}=((-1)(x-3)^2+r^2))(-x+3)


we can use point slope form
for apoint (h,k) and slope m, the equation is
y-k=m(x-h)
not sure if you want in terms of what
I'll just say for a point (h,k)
so we get
y-k=((-1)(x-3)^2+r^2))(-x+3)(x-h)
where k=2+\sqrt{-1(h-3)^2+r^2}
that's the equation of the tangent line at (h,k) for arbitrary values of r

do the plots and other stuff yourself
7 0
3 years ago
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