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3 years ago

Question 5(Multiple Choice Worth 1 points)

Mathematics

1 answer:

Snezhnost [94]3 years ago

7 0

Given:

The graph passes through the points (0,-20) and (4,10).

To find:

The equation of line that most closely represents the line depicted in the graph.

Solution:

If a line passing through two points, then the equation of line is

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

The given line passes through the points (0,-20) and (4,10). So, the equation of line is

$y-(-20)=\dfrac{10-(-20)}{4-0}(x-0)$

$y+20=\dfrac{10+20}{4}(x)$

$y+20=\dfrac{30}{4}x$

$y+20=\dfrac{15}{2}x$

Subtracting 20 from both sides, we get

$y=\dfrac{15}{2}x-20$

The function form is

$f(x)=\dfrac{15}{2}x-20$

f of x equals fifteen halves times x minus 20.

Therefore, the correct option is B.

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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.