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3 years ago

Use technology to create an appropriate model of the data.

Mathematics

2 answers:

Citrus2011 [14]3 years ago

7 0

Answer:

f(x) = 3x - 3

Step-by-step explanation:

Plug into TI-83/84 calculator.

Hit STAT

Hit EDIT

X-Values go in for L1

Y-Values go in for L2

Hit STAT

Scroll over to CALC

Hit #4 for LinReg (Finds line of best fit)

Enter all the way through

A is your slope

B is your y-intercept

Lilit [14]3 years ago

4 0

Answer:

$f(x)=3x-3$

Step-by-step explanation:

Data : (-2,-9), (0,-3), (1,0), (3,6), (5,12)

Using technology , Plot the points on the calculator .

(Refer the attached figure)

First calculate the slope of given points

$m = \frac{y_2-y_1}{x_2-x_1}$ ---A

$(x_1,y_1)=(-2,-9)$

$(x_2,y_2)=(0,-3)$

Substitute values in A

$m = \frac{-3-(-9)}{0-(-2)}$

$m = \frac{-3+9}{2}$

$m = \frac{6}{2}$

$m = 3$

$(x_1,y_1)=(1,0)$

$(x_2,y_2)=(3,6)$

Substitute values in A

$m = \frac{6-0}{3-1}$

$m = \frac{6}{2}$

$m = 3$

Since the slopes are same .

So, the the given data is a linear function.

Now to obtain the equation for data we will use two point slope form.

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

$(x_1,y_1)=(-2,-9)$

$(x_2,y_2)=(0,-3)$

Substitute values in B

$y+9= \frac{-3-(-9)}{0-(-2)}(x+2)$

$y+9= \frac{6}{2}(x+2)$

$y+9=3(x+2)$

$y+9=3x+6$

$y=3x+6-9$

$y=3x-3$

So, $f(x)=y=3x-3$

Thus Option B is true .

Hence the required equation is $f(x)=3x-3$

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3 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01

slavikrds [6]

Answer:

a) $z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

b) For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Step-by-step explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

$\hat p=\frac{174}{594}=0.293$ estimated proportion of yellow peas

$p_o=0.23$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of yellow peas is 0.23:

Null hypothesis: $p=0.23$

Alternative hypothesis: $p \neq 0.23$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>3.649)=0.00026$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) Critical value

For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

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mars1129 [50]

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3 years ago

the length of a rectangle is five times the width. if the area of the rectangle is 125 square inches, find the length and width

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Length: 25
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3 years ago