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3 years ago

15

How do i solve this chart

2 answers:

marta [7]3 years ago

3 0

Alright so filling in the chart isnt too hard. First subtract 18 from 25 which gets you 7 for no traffic citations, because 18+7=25, so it makes sense to fill in the gap for no traffic citations. 1st column done.

Shtirlitz [24]3 years ago

3 0

I give the cells on the table numbers. See image attached. There area box 1 to 9.

We count the lower part first.
Find box 8
Solve using horizontal cells
box 7 + box 8 = box 9
25 + box 8 = 50
box 8 = 50 - 25
box 8 = 25

Find box 6
Solve using horizontal cells
box 4 + box 5 = box 6
18 + 5 = box 6
box 6 = 23

Find box 1
Solve using vertical cells
box 1 + box 4 = box 7
box 1 + 18 = 25
box 1 = 25 - 18
box 1 = 7

Find box 2
Solve using vertical cells
box 2 + box 5 = box 8
box 2 + 5 = 25
box 2 = 25 - 5
box 2 = 20

Find box 3
Solve using vertical cells
box 3 + box 6 = box 9
box 3 + 23 = 50
box 3 = 50 - 23
box 3 = 27

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3 years ago

The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cyc

BaLLatris [955]

Answer:

$y(t)= 6-3cos(\dfrac{2\pi}{14}t )$

Step-by-step explanation:

The function that could model this periodic phenomenon will be of the form

$y(t) = y_0+Acos(wt)$

The tide varies between 3ft and 9ft, which means its amplitude $A$ is

$A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}$

and its midline $y_0$ is

$y_o=3+3 \\\\\boxed{y_o= 6ft}$ .

Furthermore, since at $t=0$ the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is $-cos(\omega t)$ .

The period of the full cycle is 14 hours, which means

$\omega t =2\pi$

$\omega (t+14)= 4\pi$

giving us

$\boxed{\omega = \dfrac{2\pi}{14}.}$

With all of the values of the variables in place, the function modeling the situation now becomes

$\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}$

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3 years ago