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2 years ago

Help ASAP please

Mathematics

1 answer:

r-ruslan [8.4K]2 years ago

4 0

Answer might be C hope this helps

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Answer plz. Give an explanation so I know ur not just getting points so I dont report. Thank you!

wariber [46]

Answer: A, D

Step-by-step explanation:

68 is the initial amount.

A is correct because 0.2(68) represents 20% of 68. Since you're looking for 80% of 68, you subtract 0.2(68) from 68.

D is correct because it directly takes 80% of 68, which is what the question is asking for.

6 0

2 years ago

Study the following data set.

Katen [24]

The interquartile range of the data set is 8

<h3>How to solve for the interquartile range</h3>

To do this we have to arrange in ascending order

= 8,9,9,9,10,11,13,15,17,18,22

Q2 = The median of the data is 11.

We have to find the median of the half before 11 and the median of the half after 11.

We would have

8,9,9,9,10

Q1 = 9

And 13,15,17,18,22

Q3 = 17

The interquartile range is

Q3 - Q1

= 17 - 9

= 8

The interquartile range of the data set is 8

Read more on interquartile range here:

brainly.com/question/12967898

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Correct question

Study the following data set.

{8,15,9,18,9,17,22,10,11,9,13}

What is the interquartile range of the data set?

7 0

2 years ago

What are the constants in the expression below? Check all that apply.

AlexFokin [52]

Constants are numbers alone with no variables.

The constants are: 12, -3.7, 1/3

4 0

3 years ago

Read 2 more answers

What is the answer to this equation?<br>-104 = 8x

Olegator [25]

Answer:

x=-13

Step-by-step explanation:

-104 divided by 8 to isolate x

4 0

3 years ago

Read 2 more answers

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago