15% of 20 girls who<br> Play on the team also play another sport. How many girls play another sport?

Item Value

7nadin3 [17]

Answer:

TRUE

Step-by-step explanation:

HOPE THIS HELPED!!

7 0

2 years ago

Trent went skiing with his family. His mom and dad made him put on many different layers of clothing so that he would be warm wh

mafiozo [28]

I think the answer is A. conductor

7 0

3 years ago

Read 2 more answers

1. Determine the measure of the unknown angles indicated by letters. Justify your answers with

Maurinko [17]

Answer:

hello,

Step-by-step explanation:

a)

In an isocele triangle, base's angles have the measure:

42+2a=180

2a=180-42

a=69(°)

b)

in a triangle, an external angle has for measure the sum of the angles not adjacents.

55+b=132

b=77 (°)

c)

in a quadrilater the sum of the (interior) angles is 2*180=360 degrees.

90+90+68+c=360

c=360-90-90-68

c=112 (°)

7 0

2 years ago

Round off the following numbers to two decimal places <br> 1.2.1) 0,77677<br> 1.2.2) 34,2784682

andre [41]

1) 0.78
2) 34.28
This is because the number in the 3rd decimal place for both numbers is 5 or above so it is rounded up rather than down.

8 0

3 years ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

$\sum_{n=1}^{inf}\frac{1}{n^p}$

if p>1 then series converges. In oue case we have:

$\sum_{n=1}^{inf}b_n=\frac{1}{n^4}$

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

5 0

3 years ago

15% of 20 girls who Play on the team also play another sport. How many girls play another sport?