What is the value of 1 in 58.132

What is the percent Of decrease from 50 to 21 Please anwser accurate and I give brailiest

kogti [31]

Answer:

29%

Step-by-step explanation:

To find the decrease from 50 to 21 you would subtract 50 - 21. You get 29. This is the same as 29/100. 29/100 is also 0.29 or 29%.

Hope it helps! (I'm so sorry if it's incorrect).

4 0

3 years ago

Read 2 more answers

Use the alternating series test to determine the convergence/divergence of the series

Svetach [21]

Answer:

The series converges.

Step-by-step explanation:

According to the alternating series test:

For a series ∑(-1)ⁿ aₙ or ∑(-1)ⁿ⁺¹ aₙ

If lim(n→∞) aₙ = 0

and aₙ is decreasing,

then the series converges.

aₙ = n² / (n³ + 1)

Since the power of the numerator is less than the power of the denominator, lim(n→∞) aₙ = 0.

Since n² / (n³ + 1) > (n+1)² / ((n+1)³ + 1), the series is decreasing. (We could also prove this by showing that the derivative is negative.)

Therefore, the series converges.

7 0

3 years ago

According to an airline, flights on a certain route are on time 90 90% of the time. Suppose 20 20 flights are randomly selecte

Brilliant_brown [7]

Answer:

(a) Yes, the above experiment is a binomial distribution.

(b) Probability that exactly 18 flights are on time is 28.5%.

(c) Probability that at least 18 flights are on time is 67.7%.

(d) Probability that fewer than 18 flights are on time is 32.3%.

(e) Probability that between 17 and 19 flights, inclusive, are on time is 74.5%.

Step-by-step explanation:

We are given that according to an airline, flights on a certain route are on time 90% of the time.

Suppose 20 20 flights are randomly selected and the number of on time flights is recorded.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......$

where, n = number trials (samples) taken = 20 flights

r = number of success

p = probability of success which in our question is probability that

flights on a certain route are on time, i.e; p = 0.90

Let X = Number of flights on a certain route that are on time

So, X ~ Binom(n = 20, p = 0.90)

(a) Yes, the above experiment is a binomial distribution as the probability of success is the probability in a single trial.

And also, each flight is independent of another.

(b) Probability that exactly 18 flights are on time is given by = P(X = 18)

P(X = 18) = $\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}$

= $190 \times 0.90^{18} \times 0.10^{2}$

= 0.285

Therefore, probability that exactly 18 flights are on time is 28.5%.

(c) Probability that at least 18 flights are on time is given by = P(X $\geq$ 18)

P(X $\geq$ 18) = P(X = 18) + P(X = 19) + P(X = 20)

= $\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}+\binom{20}{19} \times 0.90^{19} \times (1-0.90)^{20-19}+\binom{20}{20} \times 0.90^{20} \times (1-0.90)^{20-20}$

= $190 \times 0.90^{18} \times 0.10^{2}+20 \times 0.90^{19} \times 0.10^{1}+1 \times 0.90^{20} \times 0.10^{0}$

= 0.677

Therefore, probability that at least 18 flights are on time is 67.7%.

(d) Probability that fewer than 18 flights are on time is given by = P(X<18)

P(X < 18) = 1 - P(X $\geq$ 18)

= 1 - 0.677 = 0.323

Therefore, probability that fewer than 18 flights are on time is 32.3%.

(e) Probability that between 17 and 19 flights, inclusive, are on time is given by = P(17 $\leq$ X $\leq$ 19)

P(17 $\leq$ X $\leq$ 19) = P(X = 17) + P(X = 18) + P(X = 19)

= $\binom{20}{17} \times 0.90^{17} \times (1-0.90)^{20-17}+\binom{20}{18} \times 0.90^{18} \times (1-0.90)^{20-18}+\binom{20}{19} \times 0.90^{19} \times (1-0.90)^{20-19}$