1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
leonid [27]
3 years ago
11

Solve for x. 2x-4/10 = 2x/20

Mathematics
1 answer:
Nonamiya [84]3 years ago
7 0
<span>I think it's x=<span>419 well i could be wrong just an elementary eight year old</span></span>
You might be interested in
Who needs special prayer
kupik [55]

Answer:

me, please me

Step-by-step explanation:

I'm going through court cases because my dad raped me and so on

8 0
4 years ago
Read 2 more answers
Choose the equation that represents a line that passes through points (−1, 2) and (3, 1).
nlexa [21]

Answer:

(B) x + 4y = 7

Step-by-step explanation:

8 0
3 years ago
The first 2 terms of geometric sequnce are shown 3 and 9. What is the next term in the geometric sequence?
Bad White [126]
The next number is 27.
You can find this by finding the common ratio between 3 and 9 and multiplying 9 by the common ratio. The common ratio is 3
8 0
3 years ago
Given a population with a mean of muμequals=100100 and a variance of sigma squaredσ2equals=3636​, the central limit theorem appl
lakkis [162]

Answer:

a) \bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)

\mu_{\bar X}=100 \sigma^2_{\bar X}=1

b) P(\bar X >101)=1-P(\bar X

c) P(\bar X

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations 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".  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: P(A)+P(A') =1

Let X the random variable that represent the variable of interest on this case, and for this case we know the distribution for X is given by:  

X \sim N(\mu=100,\sigma=6)  

And let \bar X represent the sample mean, by the central limit theorem, the distribution for the sample mean is given by:  

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

a. What are the mean and variance of the sampling distribution for the sample​ means?

\bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)

\mu_{\bar X}=100 \sigma^2_{\bar X}=1.2^2=1.44

b. What is the probability that x overbarxgreater than>101

First we can to find the z score for the value of 101. And in order to do this we need to apply the formula for the z score given by:  

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

If we apply this formula to our probability we got this:  

z=\frac{101-100}{\frac{6}{\sqrt{25}}}=0.833  

And we want to find this probability:

P(\bar X >101)=1-P(\bar X

On this last step we use the complement rule.  

c. What is the probability that x bar 98less than

First we can to find the z score for the value of 98.

z=\frac{98-100}{\frac{6}{\sqrt{25}}}=-1.67  

And we want to find this probability:

P(\bar X

5 0
3 years ago
f a sampling distribution is created using samples of the amounts of weight lost by 51 people on this diet, what would be the st
galina1969 [7]

This question is incomplete, the complete question is;

A study on the latest fad diet claimed that the amounts of weight lost by all people on this diet had a mean of 24.6 pounds and a standard deviation of 8.0 pounds.

Step 2 of 2 : If a sampling distribution is created using samples of the amounts of weight lost by 51 people on this diet, what would be the standard deviation of the sampling distribution of sample means? Round to two decimal places, if necessary.

Answer:

the standard deviation of the sampling distribution of sample means is 1.12

Step-by-step explanation:

Given the data in the question,

population mean; μ = 24.6  pounds

Population standard deviation; σ = 8.0 pounds

sample size; n = 51

Now determine the standard deviation of the sampling distribution of sample means.

standard deviation of the sampling distribution of sample means is simply

⇒ population standard deviation / √sample size

= 8.0 / √51

= 8.0 / 7.141428

= 1.120224 ≈ 1.12

Therefore, the standard deviation of the sampling distribution of sample means is 1.12

5 0
3 years ago
Other questions:
  • A part time contractor at a manufacturing facility earn $11.00 per hour. He worked 4 1/4 hours on Tuesday 3 1/4 hours on Wednesd
    6·1 answer
  • {29, 29, 29, 28, 28, 27}
    11·2 answers
  • Mario is setting up a new tent during a camping trip. The tent came with 7 feet of rope. The instructions were to use 34.5 inche
    15·1 answer
  • 0.79 x 3.7 =<br> please explain your process
    8·1 answer
  • Joseph jumped off of a cliff into the ocean in Miami while vacationing with some friends. His height as a function of time is gi
    6·1 answer
  • If you need help with Math:)
    7·2 answers
  • Did I get this correct
    5·1 answer
  • Find the total cost with tip.
    9·2 answers
  • The expression (8+8+8+8+8)−(6+6+6+6+6) is equal to 10 .
    15·1 answer
  • ?? math-domain and range
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!