James answered 84% of the test questions correctly. He answered 21 questions correctly. How many questions were on the test?

6<br> Σ (12 – 4 i)<br> i = 3

AnnyKZ [126]

Answer:

- 24

Step-by-step explanation:

We have to get the value of the following sum as given.

$\sum_{3}^{6} (12-4i)$

This can be written in language as the summation of (12 -4i) where value of i ranges from 3 to 6.

$\sum_{3}^{6} (12-4i)$

= [12 - 4(3)] + [12 - 4(4)] + [12 - 4(5)] + [12 - 4(6)]

= 0 + [- 4] + [- 8] + [- 12]

= - 24 (Answer)

3 0

2 years ago

Tony spent on books to give away as prizes to his students. At the store, he bought more books than planned, spending per book.

jek_recluse [69]

Answer:

100/5-4

Step-by-step explanation:

He spend 100 total, and he spend $5 on each book. So 100/5 would give him how many books total that he bought. If he bought 4 extra you need to subtract that to find out how many books he originally wanted to buy.

8 0

3 years ago

What is y=x+2 graphed

Travka [436]

The graphing is shown below

4 0

3 years ago

Select the correct answer. Which sentence correctly describes a data set that follows a normal distribution with a standard devi

Artyom0805 [142]

Answer: Option (c) is correct.

68% of the data points lie between 10 and 18.

Step-by-step explanation: Given : a normal distribution with a standard deviation of 4 and a mean of 14

We have to choose the sentence that correctly describes a data set that follows a normal distribution with a standard deviation of 4 and a mean of 14.

Since, given 68% data.

We know mean of data lies in middle.

And standard deviation is distribute equally about the mean that is 50% of values less than the mean and 50% greater than the mean.

So, 68% of data lies

mean - standard deviation = 14 - 4 = 10

mean + standard deviation = 14 + 4 = 18

So, 68% of the data points lie between 10 and 18.

5 0

1 year ago

Read 2 more answers

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

b)

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago