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crimeas [40]
3 years ago
9

You are repairing your computer and need to loosen a screw that is fastening the top case down. You tried a 3/4 mm screwdriver,

which was too large, and a 5/8 mm screwdriver, which was too small. Which of the following screwdrivers might fit the screw?
Mathematics
1 answer:
mamaluj [8]3 years ago
6 0

Answer:

Since the screw is smaller than 3/4 mm and larger than 5/8 mm,

we can make an inequality from the given data

3/4 > x > 2.5/4

where x is the width of the screw

So since the value is greater than 2.5/4 and less than 3/4,

the approximate value will be the mean of the two values

(2.5/4) + (3/4) = 5.5/4 = 11/8 mm

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How to find derivative of a function at a given point?
seraphim [82]

Fill in the point values in the formula for the derivative.

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<u>Example</u>

y = x^2 + 3x . . . . . we want y' at (x, y) = (1, 4)

y' = 2x +3 . . . . . . . take the derivative dy/dx of the function

Fill in the value x=1 ...

y' = 2·1 +3 = 5

The value of the derivative at (x, y) = (1, 4) is 5.

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3 years ago
Find the remainder using long division. 4,598 ÷ 8 A) 2 B) 4 C) 6 D) 8
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I think it would be C, 6
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Answer: The rate of change in the y coordinate is 75. 78 - 3= 75.

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A golf course charges $16 for a package including the full 18-hole course. The course also sells buckets of
MA_775_DIABLO [31]

16x + 21y = 555

Step-by-step explanation:

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6 0
3 years ago
An individual who has automobile insurance from a certain company is randomly selected. Let X be the number of moving violations
grin007 [14]

Answer:

(X)      0        1          2          3         4

P(X)   0.17   0.23    0.27    0.24    0.09

F(x)    0.17    0.04    0.65    0.91      1      

Step-by-step explanation:  

Given that;

(X)      0        1          2          3         4

P(X)   0.17   0.23    0.27    0.24    0.09

cumulative distribution function can be calculated by;  be cumulatively up the value of p(x) with the values before it;

so

x      F(x)

0     P(X = 0) = 0.17

1       P(X = 0) + P(X = 1) = 0.17 + 0.23 = 0.4

2      P(X = 0) + P(X = 1) + P(X = 2) = 0.17 + 0.23 + 0.27 = 0.65

3      P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.17 + 0.23 + 0.27 + 0.24 = 0.91

4      P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.17 + 0.23 + 0.27 + 0.24 + 0.09 = 1

Therefore, cumulative distribution function f(x) is;

(X)      0        1          2          3         4

P(X)   0.17   0.23    0.27    0.24    0.09

F(x)    0.17    0.04    0.65    0.91      1      

8 0
2 years ago
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