Determine the range of the following graph:

Y=6/13X is a proportional relationship

Ivan

Answer:

x in (-oo:+oo)

Y = (6/13)*x // - (6/13)*x

Y-((6/13)*x) = 0

Y+(-6/13)*x = 0

Y-6/13*x = 0 // - Y

-6/13*x = -Y // : -6/13

x = -Y/(-6/13)

x = 13/6*Y

5 0

3 years ago

Translate the word phrase into a variable expression. seven less than the quotiont of x and 9

aev [14]

Quotient of x and 9= $\frac{x}{9}$

And 7 less than from it means= $\frac{x}{9} - 7$ (Answer)

8 0

3 years ago

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.