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bulgar [2K]
3 years ago
6

Write 5 x 10^6 in standard notation.

Mathematics
2 answers:
dimaraw [331]3 years ago
6 0

Answer:

So 5⋅106=5000000 or five million.

Step-by-step explanation:

34kurt3 years ago
4 0
<h2>Answer: </h2><h2>5000000</h2><h2>~</h2><h3>Since the exponent of the scientific notation is positive, move the decimal point 6 places to the right, and you'll get the answer, </h3><h3>5000000</h3><h2>~</h2>

<em>Thank you for taking the time to read this!</em>

<em>And I hope you have a good day, Loves!</em>

<em>Also if I'm right can you let me know, please?</em>

<em>Also Also if I am right can I please have Brainliest??</em>

<em>You don't need to, I'm just glad I could help. ^^</em>

<em>(If I did help-)</em>

<em>And don't forget to eat a full meal, and have at least one cup of water today! </em>

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When a scientist conducted a genetics experiments with peas, one sample of offspring consisted of 943 peas, with 717 of them hav
pshichka [43]

Using the normal approximation to the binomial distribution, it is found that:

a) 0.242 = 24.2% probability of getting 717 or more peas with red flowers.

b) Since Z < 2, 717 peas with red flowers is not significantly high.

c) Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

For each pea, there are only two possible outcomes. Either they have a red flower, or they do not. The probability of a pea having a red flower is independent of any other pea, which means that the binomial distribution is used to solve this question.

Binomial distribution:

Probability of x successes on n trials, with p probability.

Normal distribution:

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

Z = \frac{X - \mu}{\sigma}

  • 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.
  • If Z > 2, the result is considered <u>significantly high</u>.

If np \geq 10 and n(1-p) \geq 10, the binomial distribution can be approximated to the normal with:

\mu = np

\sigma = \sqrt{np(1-p)}

In this problem:

  • 943 peas, thus, n = 943
  • 3/4 probability of being red, thus p = \frac{3}{4} = 0.75.

Applying the approximation:

\mu = np = 943(0.75) = 707.25

\sigma = \sqrt{np(1-p)} = \sqrt{943(0.75)(0.25)} = 13.297

Item a:

Using continuity correction, this probability is P(X \geq 717 - 0.5) = P(X \geq 716.5), which is <u>1 subtracted by the p-value of Z when X = 716.5</u>.

Then:

Z = \frac{X - \mu}{\sigma}

Z = \frac{716.5 - 707.25}{13.297}

Z = 0.7

Z = 0.7 has a p-value of 0.758.

1 - 0.758 = 0.242

0.242 = 24.2% probability of getting 717 or more peas with red flowers.

Item b:

Since Z < 2, 717 peas with red flowers is not significantly high.

Item c:

Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

A similar problem is given at brainly.com/question/25212369

6 0
3 years ago
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