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2 years ago

Given the function f(x) =5/4 x -5 solve for the X- and y- intercepts use the interprets to graph the function .

Mathematics

1 answer:

const2013 [10]2 years ago

4 0

Answer:

y intercept = -5

x intercept = 4

Step-by-step explanation:

f(x) can be substituted with y for this question.

y = 5/4x - 5

the way to solve for x intercept is to set y to 0 and solve for x

0 = 5/4x - 5

5 = 5/4x

multiply by four on both sides

20 = 5x

divide by five on both sides

4 = x

the way to solve for the y intercept is to set x to 0 and solve for y

y = 5/4(0) - 5

y = -5

you can then graph the points (4,0) and (0,-5) to create the line for the function.

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What makes this a binomial distribution and what is the probability?

kati45 [8]

Answer:

Probability at least one car will get punctured: 0.39347

Step-by-step explanation:

B(10,000 , 0.00005)

P(X ≥ 1) = 1 - P(X = 0)

= 1 - (1 - 0.00005)^10,000

= 1 - (0.99995)^10,000

= 1 - 0.60652...

= 0.39347 (probability that at least one car will get punctured)

As you can tell P(X ≥ 1) as we have to solve for the probability that at least one car will get punctured. That is of course 1 - [ P(X = 0) ].

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3 years ago

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You spent 1/2 of your money at the theater and 1/4 at an arcade. you have $17.50 left. How much money did you originally have?

4vir4ik [10]

You basically add 17.50+17.50=35 and add 35+35=70, reason being is because the 35 is the 1/2 and 17.50 is the 1/4 so 70 is your answer

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3 years ago

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3241004551 [841]

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3 years ago

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You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling

BARSIC [14]

Answer:

(a) 15 m

(b) ground speed: 26.14 m/s (94.1 km/h); relative speed 12 m/s

Step-by-step explanation:

We can choose the frame of reference to be the trailing car. We can start counting time from the point the lead car begins deceleration. Then its position (in meters) relative to the trailing car is ...

d(t) = 25 - 5/2t² . . . . . . where 25 m is the initial distance and -5 is the acceleration in m/s².

(a) Then the distance between cars at t=2 is ...

d(2) = 25 - (5/2)·4 = 15 . . . . meters

(b) The relative velocity at 2.4 seconds is ...

d'(t) = -5t

d'(2.4) = -5(2.4) = -12.0 . . . m/s

The closure speed with the police car is 12 m/s.

We need to do a little more work to find the ground speed of the trailing car when the cars bump.

The distance between the cars at t=2.4 seconds is ...

d(2.4) = 25 -(5/2)2.4² = 10.6 . . . . meters

At the closure speed of 12 m/s, it will take an additional ...

10.6 m/(12 m/s) = 0.8833... s = 53/60 s

to close the gap. The speed of the trailing car at that point in time will be the original speed less the deceleration for 0.8833 seconds:

(110 km/h)·(1/3.6 (m/s)/(km/h))-(5 m/s²)(53/60 s)

= 26 5/36 m/s = 94.1 km/h

_____

Comment on following distance

To avoid collision, the trailing car must be trailing by at least the distance it covers in 2.4 seconds, about 73 1/3 meters.

6 0

3 years ago

A -42+(-17) b 8-(-9) c 8(-9) D E F G

Diano4ka-milaya [45]

Answer:

.A -42+(-17)

Step-by-step explanation:

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3 years ago