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hjlf
2 years ago
15

Aubrey walks 15 miles in 4 hours. At what unit rate does Aubrey walk?

Mathematics
1 answer:
Gemiola [76]2 years ago
6 0
I think it’s gonna be B.
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A car salesman sells cars with prices ranging from $5,000 to $45,000. The histogram shows the distribution of the numbers of car
MatroZZZ [7]
The distribution will exhibit symmetry. 
6 0
2 years ago
Read 2 more answers
A local pizzeria offers 15 toppings for their pizzas and you can choose any 3 of them for one fixed price. How many different ty
Shtirlitz [24]

Answer:

  455 or 680, depending

Step-by-step explanation:

If we assume the three choices are different, then there are ...

  15C3 = 15·14·13/(3·2·1) = 35·13 = 455

ways to make the pizza.

___

If two or three of the topping choices can be the same, then there are an additional ...

  2(15C2) +15C1 = 2·105 +15 = 225

ways to make the pizza, for a total of ...

  455 + 225 = 680

different types of pizza.

__

There is a factor of 2 attached to the number of choices of 2 toppings, because you can have double anchovies and tomato, or double tomato and anchovies, for example, when your choice of two toppings is anchovies and tomato.

_____

nCk = n!/(k!(n-k)!)

6 0
2 years ago
A small ball is released from rest from a point that is 40m above horizontal ground. The ball bounces on the ground and rebounds
Marina CMI [18]

Answer:

(a) 5.714s

(b) 0.625m

Step-by-step explanation:

Acceleration due to gravity=a=±9.8m/s/s (+ when falling, - when going upwards)

(a)

STAGE A: between start and first bounce

Initial speed=u=0m/s, Distance=s=40m, and Final velocity=v

v²=u²+2as

v²=0²+2(10)(40)

v²=794

v=28m/s (As the ball hits the ground)

v=u+at

28=0+9.8t_{a}

t_{a}=28/9.8=2.85714285714s

STAGE B: from the first bounce until it starts to fall

u=half the speed it hit the ground with=(28/2)=14m/s, and v=0m/s(when it starts to fall)

v=u+at

0=14-9.8t_{b}

t_{b}=14/9.8=1.42857142857s

STAGE C: between when it starts to fall after bounce 1, and bounce 2

We can assume that the time it takes to go from the ground to max height after bounce 1 is equal to the time it takes to fall from that same height to the ground. Therefore, t _{c}=1.42857142857s

The time from when the ball was released to when it hit the ground for the second time = t

t = t_{a} + t_{b} + t_{c}

t=2.85714285714+1.42857142857+1.42857142857

t=5.71428571428s≈5.714s

<u />

(b)

Before bounce 1, u=28m/s (see stage a)

After bounce 1, u= 28/2= 14m/s

After bounce 2, u= 14/2 = 7m/s

After bounce 3, u= 7/2 =3.5m/s

u=3.5m/s and v=0m/s(when it starts to fall)

v²=u²+2as

0²=3.5²+2(-9.8)s

19.6s=12.25

s=<u>0.625m (max height after third bounce)</u>

8 0
1 year ago
If a and B are the zeros of the quadratic polynomial f(x) = x2- 5x + 4 find the value of 1/a+1/b-2ab
bearhunter [10]

Answer:

<h2>-27/4</h2>

Step-by-step explanation:

Given the quadratic polynomial given as g(x) = x²- 5x + 4, the zeros of the quadratic polynomial occurs at g(x) = 0 such that x²- 5x + 4 = 0.

Factorizing the resulting equation to get the roots

x²- 5x + 4 = 0

(x²- x)-(4 x + 4) = 0

x(x-1)-4(x-1) = 0

(x-1)(x-4) = 0

x-1 = 0 and x-4 = 0

x = 1 and x = 4

Since a and b are known to be the root then we can say a = 1 and b =4

Substituting the given values into the equation  1/a+1/b-2 ab , we will have;

= 1/1 + 1/4 - 2*1*4

= 1 + 1/4 - 8

= 5/4 - 8

Find the Lowest common multiple

= (5-32)/4

= -27/4

<em>Hence the required value is -27/4</em>

5 0
2 years ago
Two fifths of the candies in a bag are mint. There are 34 mint candies.
VikaD [51]

So 2/5 are mint. that should mean if you divide 34 with 2, you will get 1/5 of the candy. SO! Divide 34 with 2 and you will get 17. The muiltiply with 5 then you will get the answer...

85 candies!!!

6 0
2 years ago
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