Here is an inequality: 7x+6/2

Which EST graph correctly represents the information on the table

garik1379 [7]

Answer:

the first one

Step-by-step explanation:

A graph showing the Earliest Start Times (EST) for project tasks is computed left to right based on the predecessor task durations. For dependent tasks, the earliest start time will be the latest of the finish times of predecessor tasks.

The first graph appears to appropriately represent the table values, using edges to represent task duration, and bubble numbers to represent start times.

The second graph does not appropriately account for duration of predecessor tasks.

The third graph seems to incorrectly compute task completion times (even if you assume that the edge/bubble number swap is acceptable).

3 0

2 years ago

Tim bought some 25 cent and some 29 cent stamps. He paid 7.60 dollars for 28 stamps. How many of each type of stamp did he buy?

Citrus2011 [14]

Answer:

25 cent stamps = 13 and 29 cent stamps = 15

Step-by-step explanation:

x = 25 cent stamps and y = 29 cent stamps

x + y = 28......x = 28 - y

0.25x + 0.29y = 7.60

0.25(28 - y) + 0.29y = 7.60

7 - 0.25y + 0.29y = 7.60

-0.25y + 0.29y = 7.60 - 7

0.04y = 0.60

y = 0.60 / 0.04

y = 15 <===== 29 cent stamps

x + y = 28

x + 15 = 28

x = 28 - 15

x = 13 <===== 25 cent stamps

lets check it...

0.25x + 0.29y = 7.60

0.25(13) + 0.29(15) = 7.60

3.25 + 4.35 = 7.60

7.60 = 7.60 (correct..it checks out)

8 0

3 years ago

Which of the following is equivalent to (5)^7/3? 5^-4<br> 5^4<br> ^7squroot5^3<br> ^3squroot5^7

kvv77 [185]

Answer:

^7 squroot5^3

Step-by-step explanation:

5x2= 10

7÷ 3= 2.5

8 0

3 years ago

Below is the graph of y= x.<br> Translate it to make it the graph of y=<br> x-1 + 4.

Oduvanchick [21]

First, to shift the graph of $y=x$ 1 unit to the right, so $y=x-1$
Second, to shift the graph of $y=x-1$ 4 units to the left, so $y=x-1+4$

<h2>Explanation:</h2>

To translate the graph of a function is part of Rigid Transformations because the basic shape of the graph is unchanged

$Let \ c \ be \ a \ positive \ real \ number. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:$

$\bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ h(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ h(x)=f(x)-c$

$\bullet \ Horizontal \ shift \ c \ units \ to \ the \ right \ \mathbf{right}: \\ h(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ left \ \mathbf{left}: \\ h(x)=f(x+c)$