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zhuklara [117]
3 years ago
11

Kyra observed that 85% of the students at a library liked reading mystery stories. If 30 students at the library did not like re

ading mystery stories, the number of students at the library who liked mystery stories was _____. (only put numeric values, no other symbols)​
Mathematics
1 answer:
Levart [38]3 years ago
3 0

Answer:

The first thing we must do is find the percentage of students who do not like to read mystery stories.

We have then:

Then, to find the number of students who like to read mystery stories, we can make the following rule of three:

15% ----------------> 30 students

85% ----------------> x

From here, we clear the value of x.

We have then:

the number of students at the library who liked mystery stories was 170

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A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

3 0
3 years ago
Determine whether the following series converges or diverges using the integral test. Be sure to verify that the integral test c
Finger [1]

By using the integral test, series converges

What is an integral test for convergence and divergence?

⇒ It is used to prove the divergence or convergence of series. This test is called the integral test, which compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.

How to know if a series is converging or diverging?

If the limit exists and is a finite number (a number less than infinity), we say the integral converges. If the limit is ±∞ or does not exist, we say the integral diverges.

let aₓ= f (x) is any function

⇒ In order for the integral test to work The function must be positive it has to be continuous and it has to be decreasing when x≥1

If the integral converges it means the series is also converging

If the integral diverges it means the series is also diverging

The sequence k^{3} /{e^{k^{4} } is clearly positive and decreases for k∈N then by the integral test,

\int\limits^\infty_1 {x^{3} /e^{x^4} dx \leq\displaystyle \sum^{\infty}_{k = 1} {x^{3} /e^{x^{4}

and

\int\limits^\infty_1 {x^{3} /e^{x^4} dx=1/4\int\limits^\infty_1 {x^{3} /e^{-u} du

⇒ 1/4 < \infty

So it comes with a finite value hence the series converges

Learn more about the integral tests here :

brainly.com/question/15394015

#SPJ1

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Factor. 6y^4+14y^3-10y^2 *
AveGali [126]

Answer:

2 y^2 (3 y^2 + 7 y - 5)

Step-by-step explanation:

Factor the following:

6 y^4 + 14 y^3 - 10 y^2

Factor 2 y^2 out of 6 y^4 + 14 y^3 - 10 y^2:

Answer: 2 y^2 (3 y^2 + 7 y - 5)

5 0
3 years ago
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