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1 year ago

HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

jekas [21]1 year ago

8 0

50 states 47 states 14 states 28 states 41 states.

Black_prince [1.1K]1 year ago

6 0

find data points or make up 5 new subjects first

linda 34 y/o, 50 states

bob 65, 47 states

sarah 12, 14 states

terry 23, 28 states

cindy 46, 41 states

plot these points along with the others then draw a line from (0,0) to the top of the x y plane passing through as many point as possible

create a table of values for this information afterward

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Twice the sum of a number and negative two is sixteen. find the number

dexar [7]

It would be 9 because 2x9=18. Then 18+-2=16.

5 0

3 years ago

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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 significantly high.

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 1 subtracted by the p-value of Z when X = 716.5.