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Hoochie [10]
3 years ago
9

Which of the following were main reasons for european exploration of the americas? select the two correct answers.

SAT
1 answer:
sattari [20]3 years ago
5 0

Answer:

There are three main reasons for European Exploration. Them being for the sake of their economy, religion and glory. They wanted to improve their economy for instance by acquiring more spices, gold, and better and faster trading routes. Also, they really believed in the need to spread their religion, Christianity.

hope it's helpful

thank you

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A recent survey indicated that the mean time spent on a music streaming service is 210 minutes per week for the population of a
Lapatulllka [165]

There are different variations in population size.  The best reason why the simulation of the sampling distribution is not approximately normal is that The sample size was not sufficiently large.

<h3>What takes place if a sample size is not big enough? </h3>
  • When a sample size taken by a person or a researcher  is not big or inadequate for the alpha level and also analyses that one have chosen to do,  it will limit the study statistical power.

Due to the above, the ability to know a statistical effect in one's sample if the effect are present in the population is greatly reduces.

See full options below

Which of the following would be the best reason why the simulation of the sampling distribution is not approximately normal?

A The samples were not selected at random.

B The sample size was not sufficiently large.

с The population distribution was approximately normal.

D The samples were selected without replacement.

E The sample means were less than the population mean.

Previous question

Learn more about population size from

brainly.com/question/1279360

6 0
2 years ago
What were three changes in the workplace in the united states during the 1990s
DENIUS [597]

Answer:

union membership increased .

Explanation:

yw heart it

3 0
2 years ago
The distribution of number of hours worked by volunteers last year at a large hospital is approximately normal with mean 80 and
Mazyrski [523]

Explanation:

Using the normal distribution, it is found that the probability that the volunteer selected will receive a certificate of merit given that the  number of hours the volunteer worked is less than 90 is closest to:

B 0.123

In a normal distribution with mean \muμ and standard deviation \sigmaσ , the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}Z=σX−μ

It 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, which is the percentile of X.

In this problem:

The mean is of 80, hence \mu = 80μ=80 .

The standard deviation is of 7, hence \sigma = 7σ=7 .

The minimum value is the 80th percentile, which means that it is X_mXm when Z has a p-value of 0.8.

The probability that the volunteer selected will receive a certificate of merit given that the number of hours the volunteer worked is less than 90 is P(X_m < X < 90)P(Xm<X<90) , which is the p-value of Z when X = 90 subtracted by the p-value of Z when X = X_mX=Xm , hence the p-value of Z when X = 90 subtracted by 0.8.

Z = \frac{X - \mu}{\sigma}Z=σX−μ

Z = \frac{90 - 80}{7}Z=790−80

Z = 1.43Z=1.43

Z = 1.43Z=1.43 has a p-value of 0.9236.

0.9236 - 0.8 = 0.1236, hence closest to 0.123, option B.

7 0
3 years ago
El diámetro de los pernos de una fabrica tiene una distribución normal con una media de 950 mm y una desviación estándar de 10 m
MrRa [10]

<em><u>ANSWER:</u></em>

0.178777

Explicación: X = diámetro de los pernos

Te piden:

P(952 < X < 957)

Ahora tienes que estandarizar para usar la tabla de la distribución

P(952 < X < 957)

P 952-950 < X-950 957-950 10 = 10 10

P(0.2 < Z <0.7)

Z~N(0,1)

P(0.2 < Z < 0.7) = (0.7) (0.2) 0.758036-0.579260 0.178777

Saludos!

6 0
3 years ago
Anyone knows this one !!
yawa3891 [41]

Answer:

B - Sentence 8

Explanation:

Paragraph is not about the mechanic.

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5 0
3 years ago
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