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Pepsi [2]
3 years ago
8

The net change for a certain stock is the dollar value change in the stock's closing price from the previous day's closing price

. The net changes of three stocks were -3, 1, and -2. Which net change has the greatest absolute value?
Mathematics
1 answer:
Natali [406]3 years ago
7 0
-3 has the greatest absolute value.
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Yep here we go again:
Shtirlitz [24]

Answer:

9 am it will be IV quadrant

11 it will be I quadrant

hope it helps u

5 0
3 years ago
Read 2 more answers
A road crew much repave a road that is 4/5 miles long. They can repave 1/20 miles each hour. How long will it take the crew to r
lutik1710 [3]

Answer:

16 hours

Step-by-step explanation:

Multiply the 4/5 by 4 to get the denominator as 20. So you do 4 x 4 which equals 16 and 5 x 4 which equals 20 so you get 16/20 miles or 16 hours.

3 0
3 years ago
The mean of five numbers is 15. four of the numbers are 3,19,8, and 32 whats the fifth number
qwelly [4]
Setup Average Equation:
_____________________
Average=Sum of our 4 numbers + unknown number of x
                ____________________________________
                                          Total Numbers
15 = 3 + 19 + 8 + 32 + x
       ______________
                      5
Cross Multiply
___________
3 + 19 + 8 + 32 +  = 15 x 5
x + 62 = 75
Subtract 62 from both sides of the equation to isolate x:
___________________________________________
x + 62 - 62 = 75 - 62
x = 13          
6 0
3 years ago
Find all the zeros if the function including multiplicity f(x)=(x+7)(x-3)^2(x-2+i)(x-2-i)
NISA [10]
Solve for x. The multiplicity of a root is the number of times the root appears.
x=-7 (multiplicity of 1)
x=3 (multiplicity of 2)
x=2-i (multiplicity of 1)
x=2+1 (multiplicity of 1)
6 0
3 years ago
Tags are attached to the left and right hind legs of a cow in a pasture. Let A1 be the event that the left tag is lost and the e
ioda

Answer:

Step-by-step explanation:

Given that tags are  attached to the left and right hind legs of a cow in a pasture. Let A1 be the event that the left tag is lost and the event that the right leg tag is lost. Suppose those two events are independent and A2 the event that the right leg tag is lost. Suppose those two events are independent and P(A1) = P(A2) = 0.3

a)  the probability that at least one leg tag is lost

= P(A_1 U A_2)\\= P(A_1) + P(A_2)-{(A_1 \bigcap A_2)\\=P(A_1) + P(A_2)-{(A_1 )P(A_2)

(since A1 and A2 are independent)

=0.3+0.3-0.09

=0.51

b) the probability that exactly one tag is lost, given that at least one tag is lost.

= P(one leg lost)/P(atleast one leg lost)

=P(atleast one leg lost)-P(Both lost)/0.51\\= \frac{0.51-0.09}{0.51} \\=14/17

c) the probability that exactly one tag is lost, given that at most one tag is lost.

4 0
3 years ago
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