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marta [7]
3 years ago
13

Hamid's soccer game will start at 11:00 A.M., but the players must arrive at the field three-quarters of an hour early to warm u

p. The game must end by 1:15 P.M.
Hamid says he has to be at the field at 9:45 A.M. Is Hamid correct? Explain.

My brother doesn't understand things like this when I explain them to him.
Mathematics
1 answer:
Minchanka [31]3 years ago
4 0
If the game will start at 11:00 A.M., but the players must arrive at the field three-quarters of an hour early to warm up, it refers to 8:45 a.m. Why? If we start to count in 11 backward and start to trace the three-quarters, it shows that 10:45, 9:45, and 8:45 are the three-quarters. So Hamid statement that he has to be at the field at 9:45 A.M is not correct.  
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madison and centerville are 6 cm apart on a map that has a scale of 1 cm : 11km. how far apart are the real cities
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Suppose that a box contains 7 cameras and that 5 of them are defective. A sample of 2 cameras is selected at random without repl
irinina [24]

Answer:

E(x) = 1.43 (Approx)

Step-by-step explanation:

Given:

Total number of camera = 7

Defective camera = 5

Sample selected = 2

Computation:

when x = 0

P(x=0) = 2/7 × 1/6 = 2/42

P(x=1) = [2/7 × 5/6] + [5/7 × 2/6] = 20/42

P(x=2) = 5/7 × 4/6 = 20/42

So,

E(x) = [0×2/42] + [1×20/42] + [2×20/42]

E(x) = 1.43 (Approx)

8 0
3 years ago
Can someone please help
andre [41]

Answer:

put it in the catrulater

Step-by-step explanation:

3 0
3 years ago
PLZ SHOW ALL WORK
lara31 [8.8K]

Part I
We have the size of the sheet of cardboard and we'll use the variable "x" to represent the length of the cuts. For any given cut, the available distance is reduced by twice the length of the cut. So we can create the following equations for length, width, and height.
width:  w = 12 - 2x
length: l = 18 - 2x
height: h = x

Part II
v = l * w * h
v = (18 - 2x)(12 - 2x)x
v = (216 - 36x - 24x + 4x^2)x
v = (216 - 60x + 4x^2)x
v = 216x - 60x^2 + 4x^3
v = 4x^3 - 60x^2 + 216x

Part III
The length of the cut has to be greater than 0 and less than half the length of the smallest dimension of the cardboard (after all, there has to be something left over after cutting out the corners). So 0 < x < 6

Let's try to figure out an x that gives a volume of 224 in^3. Since this is high school math, it's unlikely that you've been taught how to handle cubic equations, so let's instead look at integer values of x. If we use a value of 1, we get a volume of:
v = 4x^3 - 60x^2 + 216x
v = 4*1^3 - 60*1^2 + 216*1
v = 4*1 - 60*1 + 216
v = 4 - 60 + 216
v = 160

Too small, so let's try 2.
v = 4x^3 - 60x^2 + 216x
v = 4*2^3 - 60*2^2 + 216*2
v = 4*8 - 60*4 + 216*2
v = 32 - 240 + 432
v = 224

And that's the desired volume.
So let's choose a value of x=2.
Reason?
It meets the inequality of 0 < x < 6 and it also gives the desired volume of 224 cubic inches.
3 0
3 years ago
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