Y = -2.8x +69.4
Let y represent units of inventory, and x represent days since the last replenishment. We are given points (x, y) = (3, 61) and (13, 33). The line through these points can be described using the 2-point form of the equation of a line:
... y -y1 = (y2-y1)/(x2 -x1)(x -x1)
Filling in the given point values, we have ...
... y -61 = (33 -61)/(13 -3)(x -3)
Simplifying and adding 61, we get ...
... y = -2.8x +69.4
Answer:
pic1=56-28x
pic2=3cd+12c
pic3=s−t
pic4= the last answer
pic5= the third answer
Step-by-step explanation:
i did the math so just trust me
Answer:
m=21
Step-by-step explanation:
based on the dilation from 'ABC' to 'QRS' (which is 3) m should =21
hope this helps! :)
Answer:
Step-by-step explanation:
a) We have 15! as the product of 1 to 15 natural numbers. Since 17 is prime there will be no factor common to these
By actual division we find
15! (mod 17) =16
From this we deduce
even 16! mod 17 = 16 = -1
According to Wilson theorem
(17-1)! = -1 mod 17
Thus verified 17 is prime
Hence 15! (mod 17) =-1=16
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b) 2(26!) is divided by 29
Since 29 is prime
(29-1)! = -1 mod 29
28! = -1 mod 29 = 28
When divided this gives 25 as remainder
#1
The uniforms are numbered 0, 1, 2, ..., 99. That's 100 numbers. Half of them are odd and half of them are even. So the probability that any one of the uniforms is odd is 1/2 just like the probability that any one uniform is even is 1/2.
(a) The numbers on the uniforms are independent of one another. That is, the number of her cross-country uniform does not in any way determine the number on her basketball uniform and vice versa. This means that we can find the probability that each is odd and multiply these together using what is called the counting principle. The probability that all are odd is:
(1/2)(1/2)(1/2)=1/8
(b) This is done the same way we did part (a). Since the probability of any one uniform being odd is the same as it being even (1/2), the answer here is the same: (1/2)(1/2)(1/2)=1/8
(c) This problem differs from that in (a) and (b). There is only one way for all three uniforms to be odd numbers: (odd, odd, odd) or all even (even, even, even). However, there are multiple ways for the uniforms to be two odd and one even. If the uniforms are listed in order: cross-country, basketball, softball we can get exactly one even in any of three ways:
even, odd, odd
odd, even, odd
odd, odd, even
The probability for any one of these possibilities is (1/2)(1/2)(1/2)=1/8 but since there are three way the probability that we get even exactly once is equal to (3)(1/8) = 3/8
