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3 years ago

Graph the line for this problem: y=3x+6

Mathematics

2 answers:

sertanlavr [38]3 years ago

5 0

$y=3x+6\\\\for\ x=0\to y=3\cdot0+6=6\to A(0;\ 6)\\\\for\ x=-2\to y=3\cdot(-2)+6=-6+6=0\to B(-2;\ 0)$

Zigmanuir [339]3 years ago

5 0

$y=3x+6 \\ \\x=0 \\ \\y=3*0+6\\ \\y=6 \\ \\A(0,6)$

$y=0 \\ \\0=3x+6\\ \\3x=-6/:3\\ \\x=-2 \\ \\B(-2,0)$

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4/25, 13%, 0.28, 7%, 21/100, 0.15 least to greatest

Katarina [22]

7%, 13%, 0.15, 4/25, 21/100, 0.28

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2 years ago

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3 years ago

Given rs || qp, what is the value of x?

myrzilka [38]

In the given figure,

Angle 108° + 12x = 180°

because they for pair of co- interior angles which are supplementary.

so,

$12x + 108\degree = 180\degree$

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2 years ago

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viva [34]

Answer:

The critical value of z for 99% confidence interval is 2.5760.

The 99% confidence interval for population mean number of lightning strike is (7.83 mn, 8.37 mn).

Step-by-step explanation:

Let X = number of lightning strikes on each day.

A random sample of n = 23 days is selected to observe the number of lightning strikes on each day.

The random variable X has a sample mean of, $\bar x=8.1\ mn$ and the population standard deviation, $\sigma=0.51\ mn$ .

The (1 - α)% confidence interval for population mean μ is:

$CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

Compute the critical value of z for 99% confidence interval is:

$z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.5760$

*Use a z-table.

The critical value of z for 99% confidence interval is 2.5760.

Compute the 99% confidence interval for population mean number of lightning strike as follows:

$CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\=8.1\pm2.5760\times\frac{0.51}{\sqrt{23}}\\=8.1\pm0.2738\\=(7.8262, 8.3738)\\\approx(7.83, 8.37)$

Thus, the 99% confidence interval for population mean number of lightning strike is (7.83 mn, 8.37 mn).

The 99% confidence interval for population mean number of lightning strike implies that the true mean number of lightning strikes lies in the interval (7.83 mn, 8.37 mn) with 0.99 probability.

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2 years ago

A researcher planned a study in which a crucial step was offering participants a food reward. It was important that three food r

lbvjy [14]

Answer:

Step-by-step explanation:

Hello!

A pilot study was conducted to test if three food rewards are equally appealing to the participants.

Of 60 participants surveyed:

16 preferred cupcakes (CC)

26 preferred candy bars (CB)

18 preferred dried apricots (DA)

If the three types of food are equally appealing for the participants, you'd expect that their proportions will be equal: P(CC)=P(CB)=P(DA)= 1/3

The objective of this pilot study is to test if the observed frequencies follow a theoretical model/ distribution. To analyze this, you have to apply a Chi Square Goodness to Fit test. $X^2=sum \frac{(O_i-E_i)^2}{E_i} ~~X^2_{k-1}$ Where k= number of categories of the variable.

For this example the statistical hypotheses are:

H₀: P(CC)=P(CB)=P(DA)= 1/3

H₁: At least one of the proportions isn't equal to the others.

To calculate the expected frequencies for each category you have to use the formula: $E_i= n* P_i$ where Pi represents the theoretical proportion for the i category, stated in the null hypothesis.

$E_{CC}= n* P(CC)= 60*1/3= 20$

$E_{CB}= n*P(CB)= 60*1/3= 20$

$E_{DA}= n* P(DA)= 60* 1/3= 20$

The cutoff or critical value indicates the beginning of the rejection region for the hypothesis test. For the Chi-Square tests, the rejection region is always one-tailed to the right, meaning that you'll reject the null hypothesis if the value of the statistic is big. For the goodness to fit test you have k-1 degrees of freedom, so the critical value will be:

Assuming α: 0.05

$X^2_{k-1;1-\alpha /2}= X^2_{2;0.975}= 7.378$

The rejection region is then X²₂ ≥ 7.378

$X^2_{H_0}= \frac{(O_{CC}-E_{CC})^2}{E_{CC}} + \frac{(O_{CB}-E_{CB})^2}{E_{CB}} + \frac{(O_{DA}-E_{DA})^2}{E_{DA}} = \frac{(16-20)^2}{20} +\frac{(26-20)^2}{20} +\frac{(18-20)^2}{20}= \frac{14}{5}= 2.8$

I hope this helps!

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2 years ago