MATHEMATICS COLLEGE

A certain social site has grown rapidly, Accerding to one repart, about 6 million users were added each day, How many users were added, on average, each second? (Round your answer to the nearest integer.) users/second elook

Answers

Answer 1

Answer:

Answer:

69 users.

Step-by-step explanation:

First, we need to calculate the amount of seconds in one day.

a) Each hour has 60 minutes and each minute has 60 seconds.

Therefore, one hour has (60)(60) = 3600 seconds.

b) Since the day has 24 hours, we are going to multiply the total amount of seconds per hour by 24.

3600 (24) = 86400 seconds per day.

c) Now that we have the amount of seconds per day, we are going to divide the total amount of users added each day by the amount of seconds per day.

6000000/86400 = 69.44 users per second.

Therefore, 69 users are added, on average, each second.

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Find the greatest factor of 6w^3 and 10a^4

Answers

Answer:2

Step-by-step explanation:

Greatest factor of 6w^4 10a^4

6w^4=2x3xw^4

10a^4=2x5xa^4

There greatest common factor is 2

Complete the Ratio Table

Answers

3 beets= 1.5
4 beets= 2
6 beets= 3
i’m pretty sure it’s just half of what the quantity is sorry if i’m wrong

The picture shows a carousel. The carousel spins while the horses move up and down. A student determines toeach horse completes one up-down cycle in 30 seconds. The student rides a horse on the carousel for 2 minute
which equals 120 seconds.
raal

Answers

Answer:

4 times

Step-by-step explanation:

Note,

60 seconds = 1 minute
120 seconds = 2 minutes (60*2).

Since the student rode a horse for 120 seconds, to find the number of times the student riding the carousel completed the up-down cycle, we divide it by 30 seconds (which is the duration for the horse to complete one up-down cycle) because the student would be having the same duration as the horse because he was on top the horse.

Hence, we have;

120/30 = 4

In tilapia, an important freshwater food fish from Africa, the males actively court females. They have more incentive to court a female who has already laid all of her eggs, but can they tell the difference? an experiment was done to measure the male tilapia's response to the smell of female fish. Water containing feces from females that were either pre-ovulatory (they still had eggs) or post-ovulatory (they had already laid their eggs) was washed over the gills of males hooked up to an electro-olfactogram machine which measured when the senses of the males were excited. The amplitude of the electro-olfactogram was used as a measure of the excitability of the males in the two different circumstances. Six males were exposed to the scent of pre-ovulatory females; their readings average 1.51 with a standard deviation of .25. Six different males were exposed to post-ovulatory females; their average readings of 0.87 with standard deviation is .31. Assume that the electro-olfactogram readings were approximately normally distributed within the groups.(A) test for a difference in the excitability of the males with exposure to these two types of females
(B) what is the estimated average difference in electro-olfactogram readings between the two groups? What is the 95% confidnece limit for the difference between population means?

Answers

Answer:

a) $t=\frac{1.51-0.87}{\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}}=3.936$

"=T.INV(1-0.025,10)", and we got $t_(critical)=\pm 2.28$

Statistical decision

Since our calculated value is higher than our critical value, $z_(calc)=3.936>2.28=t_(critical)$ , we have enough evidence to reject the null hypothesis at 5% of significance.

b) $(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}$

The degrees of freedom are given:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

$(1.51 -0.87) - 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 0.269$

$(1.51 -0.87) + 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 1.010$

Step-by-step explanation:

Part a

Data given and notation

$\bar X_(1)=1.51$ represent the mean for scent of pre ovulatory

$\bar X_(2)=0.87$ represent the mean for post ovolatory

$s_(1)=0.25$ represent the sample standard deviation for preovulatory

$s_(2)=0.31$ represent the sample standard deviation for postovulatory

$n_(1)=6$ sample size for the group preovulatory

$n_(2)=6$ sample size for the group postovulatory

z would represent the statistic (variable of interest)

$p_v$ represent the p value

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean's are different, the system of hypothesis would be:

H0: $\mu_(1) = \mu_(2)$

H1: $\mu_(1) \neq \mu_(2)$

If we analyze the size for the samples both are lower than 30, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_(1)-\bar X_(2)}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We have all in order to replace in formula (1) like this:

$t=\frac{1.51-0.87}{\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}}=3.936$

Find the critical value

We find the degrees of freedom:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

In order to find the critical value we need to take in count that we are conducting a two tailed test, so we are looking for thwo values on the t distribution with df =10 that accumulates 0.025 of the area on each tail. We can us excel or a table to find it, for example the code in Excel is:

"=T.INV(1-0.025,10)", and we got $t_(critical)=\pm 2.28$

Statistical decision

Since our calculated value is higher than our critical value, $z_(calc)=3.936>2.28=t_(critical)$ , we have enough evidence to reject the null hypothesis at 5% of significance.

Part b

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}$

The degrees of freedom are given:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

$(1.51 -0.87) - 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 0.269$

$(1.51 -0.87) + 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 1.010$

The hypotenuse of the triangle shown below is 12 inches. What is the lengthof a side in inches?

Answers

Can you send a picture of the triangle ?

What time will it be in 3/4 hour

Answers

3/4 of an hour is equal to 45 minutes.

To find 3/4 of an hour, we need to multiply the value of one hour by 3/4.

One hour is equal to 60 minutes.

To calculate 3/4 of an hour:

= 3/4 x 60 minutes

= (3/4) x 60

= 45 minutes.

Therefore, 3/4 of an hour is equal to 45 minutes.

Learn more about Fraction here:

brainly.com/question/10354322

#SPJ6

Answer:

45 minutes

Step-by-step explanation:

so 1 hr= 60 minutes so you do 3/4 x 60min which is 45 minutes. Feel Free to mark Brainliest and have a great day!

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