30 points.. Please Help! I don't understand what to do with this question...a bit confused.

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

5 0

Eh, here's my best try for a table:

Like tv don't like tv total
Like reading: 70 10 80
Don't like read. 20 30 50 
total 90 40 130

The percentage of students that like tv as well as reading is- 70 divided by 130 which gives you .538 which in percentage is 53.8%.

The probability of a student that does not prefer to watch tv as also does not like to spend their time reading is 30 over 40 which can be simplified or reduced to 3 over 4. [30/40 reduced to 3/4]

Hope this somehow helped you :)

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Ill give brainiest

mario62 [17]

Answer:

I got one

Step-by-step explanation:

C.converse of the corresponding angles postulate

∠1 = 58° ( alternate angles )

HOPE THIS HELPS:)

STAY SAFE

∠1 and ∠2 form a straight angle and are supplementary, thus

∠2 = 180° - 58° = 122°

8 0

2 years ago

Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the co

jeka57 [31]

Answer:

<h2>A. The series CONVERGES</h2>

Step-by-step explanation:

If $\sum a_n$ is a series, for the series to converge/diverge according to ratio test, the following conditions must be met.

$\lim_{n \to \infty} |\frac{a_n_+_1}{a_n}| = \rho$

If $\rho$ < 1, the series converges absolutely

If $\rho > 1$ , the series diverges

If $\rho = 1$ , the test fails.

Given the series $\sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}$

To test for convergence or divergence using ratio test, we will use the condition above.

$a_n = \frac{n^2}{5^n} \\a_n_+_1 = \frac{(n+1)^2}{5^{n+1}}$

$\frac{a_n_+_1}{a_n} = \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}\$

$\frac{a_n_+_1}{a_n} = {{\frac{(n^2+2n+1)}{5^n*5^1}} * \frac{5^n}{n^2}\\$

aₙ₊₁/aₙ =

$\lim_{n \to \infty} |\frac{ n^2+2n+1}{5n^2}| \\\\Dividing\ through\ by \ n^2\\\\\lim_{n \to \infty} |\frac{ n^2/n^2+2n/n^2+1/n^2}{5n^2/n^2}|\\\\\lim_{n \to \infty} |\frac{1+2/n+1/n^2}{5}|\\\\$