Please help me! No links

Can someone please help me?

professor190 [17]

Answer:

correct 34324234324324

Step-by-step explanation:

8 0

3 years ago

Plzzzzz help,.............

mafiozo [28]

With what? Maybe I can help

5 0

3 years ago

objective: central limit theorem assumptions. the factor(s) to be considered when assessing if the central theorem holds is/are

Eddi Din [679]

Answer:

Sample size

Step-by-step explanation:

Central Limit Theorem states that population with mean and standard deviation and if the sample size is large then the distribution of sample mean will be will be normally distributed. The central limit theorem holds assumptions that the factors to be considered when assessing central limit theorem is sample size.

8 0

3 years ago

The half-life of the radioactive isotope gallium-70 is 20 minutes. The initial amount of gallium-70 is 25 milligrams. When will

stiv31 [10]

Answer:

Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is half-life, which is the amount of time it takes for one-half of a radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by coTnditions and is independent of the initial amount of that isotope.

6 0

3 years ago

Read 2 more answers

A production process is checked periodically by a quality control inspector. the inspector selects simple random samples of 30 f

777dan777 [17]

Answer:

Population Mean = 2.0

Population Standard deviation = 0.03

Step-by-step explanation:

We are given that the inspector selects simple random samples of 30 finished products and computes the sample mean product weight.

Also, test results over a long period of time show that 5% of the values are over 2.1 pounds and 5% are under 1.9 pounds.

Now, mean of the population is given the average of two extreme boundaries because mean lies exactly in the middle of the distribution.

So, Mean, $\mu$ = $\frac{1.9+2.1}{2}$ = 2.0

Therefore, mean for the population of products produced with this process is 2.

Since, we are given that 5% of the values are under 1.9 pounds so we will calculate the z score value corresponding to a probability of 5% i.e.

z = -1.6449 {from z % table}

We know that z formula is given by ;

$Z = \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

-1.6449 = $\frac{1.9 - 2.0}{\frac{\sigma}{\sqrt{n} } }$ ⇒ $\frac{\sigma}{\sqrt{n} } = \frac{-0.1}{-1.6449}$

⇒ $\sigma =$ 0.0608 * $\sqrt{30}$ {as sample size is given 30}

⇒ $\sigma$ = 0.03 .

Therefore, Standard deviation for the population of products produced with this process is 0.0333.

7 0

3 years ago