All categories

1 year ago

3.82e+7 in standard form shown in calculator display

Mathematics

2 answers:

a_sh-v [17]1 year ago

5 0

The answer is 3.82*10^7.

This is because in the calculator when the number is quite big enough or small enough it is written in a standard form which is 3.82*e^7 as the calculator is not able to write in the power of 10.

So whenever it is e to the power something it means it is 10 to the power something. In this case it was 3.82*e^7 which means 3.82*10^7. Which means 38200000.

ser-zykov [4K]1 year ago

3 0

E+7 means 10^+7
basically move the decimal place 7 placs to right so it would be aprox
38200000.0

You might be interested in

Logan wants to move to a new city. He gathered graphs of temperatures for two different cities. Which statements about the data

eduard

Answer:

we need the data to answer the question

3 0

2 years ago

A fitted multiple regression equation is Y = 28 + 5X1 - 4X2 + 7X3 + 2X4. When X1 increases 2 units and X2 increases 2 units as w

ad-work [718]

Answer:

A. Increase by 2

Step-by-step explanation:

Given that a fitted multiple regression equation is

$Y = 28 + 5X_1 -4X_2 + 7X_3 + 2X_4$

This is a multiple regression line with dependent variable y and independent variables x1, x2, x3 and x4

The coefficients of independent variables represent the slope.

In other words the coefficients represent the rate of change of y when xi is changed by 1 unit.

Given that x3 and x4 remain unchanged and x1 increases by 2 and x2 by 2 units

Since slope of x1 is 5, we find for one unit change in x1 we can have 5 units change in y

i.e. for 2 units change in x1, we expect 10 units change in Y

Similarly for 2 units change in x2, we expect -2(4) units change in Y

Put together we have

$10-8 =2$ change in y

Since positive 2, there is an increase by 2

A. Increase by 2

5 0

2 years ago

A randomized study compared two different systems for tracking baggage at an airport. The treatment group using System A reporte

Natasha2012 [34]

Answer:

D. The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

Step-by-step explanation:

The treatment group using System A reported a mean of 18.5 lost bags per day. The treatment group using System B reported a mean of 16.6 lost bags per day.

The best conclusion that can be made is - The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

As we know, in statistics, nothing happens by chance. So, this option is correct.

8 0

2 years ago

Read 2 more answers

"Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad lengt

Westkost [7]

Answer:

a) $\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})$

b) $Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5$

c) $P(\bar X >31.25)=0.006=0.6\%$

d) $P(\bar X >28.25)=0.9997=99.97\%$

e) $P(28.25$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =30,\sigma =2)$

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})$

b. What is the standard error of the mean time?

The standard error is given by this formula:

$Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5$

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(\bar X >31.25)=1-P(\bar X$

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(\bar X >28.25)=1-P(\bar X$

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

$P(28.25$

3 0

2 years ago

Carly commutes to work, and her commute time is dependent on the weather. When the weather is good, the distribution of her comm

Stels [109]

We are given with
x1 = 20 min
s1 = 2 min

x2 = 30 min
s2 = 4 min

p = 0.9
Condition (x > 25)

We need to get the t-value between the two means and comparing it wit the t-value for the time of 25 minutes given that there is a 90% probability that the weather will be good. Simply use the t-test formula and use the t-test table to get the probability.

8 0

2 years ago

Read 2 more answers