All categories

3 years ago

⚠️⚠️⚠️HELP HELP HELP IMAGE ATTACHED EXTRA POINTS

Mathematics

1 answer:

Licemer1 [7]3 years ago

3 0

Answer:

It will be 0.123 (f)t

Step-by-step explanation:

Multiplying the plan by the patient list gives the usage for the upcoming days: [3 6 9 12 15]. Everyday multiplication (3 x 4) means using the plan with a single day of patients: [3] * [4] = [12].

Intuition For Convolution

Let's say the disease mutates and requires a multi-day treatment. You create a new plan: Plan: [3 2 1]

That means 3 units of the cure on the first day, 2 on the second, and 1 on the third. Ok. Given the same patient schedule ( [1 2 3 4 5]), what's our medicine usage each day?

You might be interested in

Please help me, Thank you :)

aliina [53]

Answer:

I am pretty sure its the first one

7 0

3 years ago

Read 2 more answers

Which relation is a function?

IceJOKER [234]

Answer:

The bottom right option

Step-by-step explanation:

Functions need each input to map to exactly one output. Graphically speaking, this means that any two points on the function cannot have the same x-coordinate.

Given this, the bottom right option is the only one that fits these circumstances.

4 0

2 years ago

Chuck drops a rock from a height of 70 m. How long will it take the rock to hit the ground? Use the formula h(t)=−4.9 t 2 + v o

Pachacha [2.7K]

That's the best I could do. I hope it helped

7 0

3 years ago

Read 2 more answers

1675 at 4.6% for 4 years What is the balance

Yuri [45]

Amount of Interest (I) = P * R * T /100
I = 1675 * 4.6 * 4 /100
I = 308.20

Balance = Initial amount + Interest amount
B = 1675 + 308.20 = 1983.20

In short, Your Answer would be $1983.20

Hope this helps!

4 0

3 years ago

Consider the following ordered data. 6 9 9 10 11 11 12 13 14 (a) Find the low, Q1, median, Q3, and high. low Q1 median Q3 high (

IrinaVladis [17]

Answer:

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = 3.5

Step-by-step explanation:

Given that:

Consider the following ordered data. 6 9 9 10 11 11 12 13 14

From the above dataset, the highest value = 14 and the lowest value = 6

The median is the middle number = 11

For Q1, i.e the median of the lower half

we have the ordered data = 6, 9, 9, 10

here , we have to values as the middle number , n order to determine the median, the mean will be the mean average of the two middle numbers.

i.e

median = $\dfrac{9+9}{2}$

median = $\dfrac{18}{2}$

median = 9

Q3, i.e median of the upper half

we have the ordered data = 11 12 13 14

The same use case is applicable here.

Median = $\dfrac{12+13}{2}$

Median = $\dfrac{25}{2}$

Median = 12.5

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = Q3 - Q1

The interquartile range = 12.5 - 9

The interquartile range = 3.5

7 0

3 years ago