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Maksim231197 [3]
3 years ago
5

For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

equally Among the 4 fifth grade classes in the school what fraction of the whole athletic field is reserved for each fifth class
Mathematics
1 answer:
Makovka662 [10]3 years ago
8 0

Answer: 1/24

Step-by-step explanation:

We have that 1/6 of the total field is reserved for the fifth-grade classes.

We have 4 fifth-grade classees.

Then each one of those 4 classes, has (1/4) of the reserved space.

This is:

space for each fifth-grade class = (1/4)*(1/6) = 1/24

So each fifth-grade class has a 1/24 of the total field.

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Irina-Kira [14]
The best batch size is 1053.

Since 5% have failed, this means 95% have passed the test.

95% = 95/100 = 0.95

We can set up an equation to answer this:
0.95x = 1000

Divide both sides by 0.95:
0.95x/0.95 = 1000/0.95
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7 0
3 years ago
Can u help me ASAP. i need to know how to do it step by step​
12345 [234]

Answer:

<em>19</em>

Step-by-step explanation:

f(x) =4x^3-8x^2+ax+b has a factor 2x-1 and

when divided by x+2, remainder is 20.

To find:Remainder when divided by (x-1)  ?

Solution:

2x-1 is a factor

2x - 1 = 0 \Rightarrow x = \frac{1}{2} when we put this value of x to the function, it will become 0.

i.e.

\Rightarrow f(\dfrac{1}{2}) =0 =4\times (\dfrac{1}2)^3-8(\dfrac{1}2)^2+\dfrac{a}{2}+b=0\\\Rightarrow \dfrac{1}{2}-2+\dfrac{a}{2}+b=0\\\Rightarrow 1-4+a+2b=0\\\Rightarrow a +2b=3 ......(1)

Remainder is 20 when f(x) is divided by x+2

i.e.

f(-2) =20

\Rightarrow f(-2) =20 =4\times (-2)^3-8(-2)^2-2a+b=20\\\Rightarrow -32-32-2a+b=20\\\Rightarrow -2a+b=84 ...... (2)

Solving (1) and (2), Multiply equation (1) by 2 and adding to (2):

5b=6+84\\\Rightarrow b = \dfrac{90}{5} = \bold{18}

By equation (1):

a+2(18) = 3\\\Rightarrow a = -33

So, the equation becomes:

f(x) =4x^3-8x^2-33x+18

\Rightarrow f(1) = 4(1) -8 (1) -33(1) +18 = \bold{19}

So, when divided by (x-1), remainder will be <em>19</em>.

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3 years ago
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Answer:

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There are several ways you can list all the possible combinations. A couple of my favorite are a) use a binary counting sequence; b) use a gray code counting sequence.

Using the first method, the binary numbers 000 to 111 can be listed in numerical order as 000, <em>001</em>, 010, <em>011</em>, 100, 101, 110, 111. Letting 0=P and 1=F, the ones missing from your list are the ones in italics in my list.

Using the second method, we change the right-most character, then the middle one, and finally the left-most character so there is one change at a time: 000, <em>001</em>, <em>011</em>, 010, 110, 111, 101, 100.

After you have a list of all possible combinations, it is a simple matter to compare the given list to the list of possibilities to see which are missing.

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3 years ago
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Answer:

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10 points, 2 times 4 equals 10 points.


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