All categories

4 years ago

Which of these is a geometric sequence?

Mathematics

2 answers:

lilavasa [31]4 years ago

5 0

Answer:

i think the answer to this is c

Step-by-step explanation:

Eduardwww [97]4 years ago

5 0

Answer:

It's C.

Step-by-step explanation:

A geometric sequence has a common ratio (r) where each term after the first is obtained by multiplying the previous one by r.

Checking through the list we see it is C because:

Second term 15 = first term * 5 = 3(5) = 15.

Third term = 15*5 = 75,

4th term = 75*5 = 375 and so on.

The common ratio is 5.

You might be interested in

Catron evaluates the expression (negative 9) (2 and two-fifths) using the steps below.

Vlad [161]

Answer:

Catron's error is

"She did not follow order of operations"

Step-by-step explanation:

Catron evaluates the expression (negative 9) (2 and two-fifths)

That expression can be written as below

$(-9)(2\frac{2}{5})$

Catron's error is

"She did not follow order of operations"

The corrected steps are

Step1: Given expression is $(-9)(2\frac{2}{5})$

Step2: Convert mixed fraction into improper fraction

$(-9)(2\frac{2}{5})=(-9)(\frac{12}{5})$

Step3: Multiplying the terms

$(-9)(2\frac{2}{5})=-7\frac{-108}{5}$

Therefore solution $(-9)(2\frac{2}{5})=-7\frac{-108}{5}$

6 0

3 years ago

Read 2 more answers

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0

4 years ago

A fishing boat lies 200 m due south of a large tree on the shoreline and 300 m southwest of the dock. The shoreline runs East to

Lunna [17]

Tree...........................dock
.
.
.
.
.
boat

use Pythagorean Theorem to satisfy...a^2 + b^2 = c^2

200^2 + b^2 = 300^2
b^2 = 300^2 - 200^2
b^2 = 50000 (get square root of each)
b = 223.606 (rounded to the nearest tenth.... = 223.6

4 0

3 years ago

Read 2 more answers

A video game store was getting rid of old games,selling them 7 for 100.66. .If they sold 2 games ,how much money would they have

Pani-rosa [81]

Answer: They would make $28.76.

Explanation: If 7 video games are worth $100.66, when you divide 100.66 by 7(100.66/7) you get the amount of money that 1 video game would be worth which is $14.38. Multiply 14.38 by 2(the number of video games they sold) and you get the answer of 28.76(the total money they made).

4 0

3 years ago

I need to know the answer to 50.5/17 for my math homework

ddd [48]

Answer:2.97058823529

Step-by-step explanation:

8 0

3 years ago