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3 years ago

Can someone help me answer this please

Mathematics

1 answer:

Black_prince [1.1K]3 years ago

5 0

Answer:

(x-2)^2 +(y+3)^2 = 16

Step-by-step explanation:

The equation of a circle can be written as

(x-h)^2 +(y-k)^2 = r^2

where (h,k) is the center and r is the radius

(x-2)^2 +(y- -3)^2 = 4^2

(x-2)^2 +(y+3)^2 = 16

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What is the slope of the line through (2, -1) and (-2, -3)?

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$\text{Given that,}\\\\\\(x_1,y_1) = (2,-1)~~\text{and}~~ (x_2,y_2) = (-2,-3)\\\\\\\text{Slope,}~ m =\dfrac{y_2-y_1}{x_2 -x_1} = \dfrac{-3+1}{-2-2} = \dfrac{-2}{-4} = \dfrac 12$

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2 years ago

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An article reports the following data on yield (y), mean temperature over the period between date of coming into hops and date o

skelet666 [1.2K]

Answer:

x1=c(16.7,17.4,18.4,16.8,18.9,17.1,17.3,18.2,21.3,21.2,20.7,18.5)

x2=c(30,42,47,47,43,41,48,44,43,50,56,60)

y=c(210,110,103,103,91,76,73,70,68,53,45,31)

mod=lm(y~x1+x2)

summary(mod)

R output: Call:

lm(formula = y ~ x1 + x2)

Residuals:

Min 1Q Median 3Q Max

-41.730 -12.174 0.791 12.374 40.093

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 415.113 82.517 5.031 0.000709 ***

x1 -6.593 4.859 -1.357 0.207913

x2 -4.504 1.071 -4.204 0.002292 **

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 24.45 on 9 degrees of freedom

Multiple R-squared: 0.768, Adjusted R-squared: 0.7164

F-statistic: 14.9 on 2 and 9 DF, p-value: 0.001395

a). y=415.113 +(-6.593)x1 +(-4.504)x2

b). s=24.45

c). y =415.113 +(-6.593)*21.3 +(-4.504)*43 =81.0101

residual =68-81.0101 = -13.0101

d). F=14.9

P=0.0014

There is convincing evidence at least one of the explanatory variables is significant predictor of the response.

e). newdata=data.frame(x1=21.3, x2=43)

# confidence interval

predict(mod, newdata, interval="confidence")

#prediction interval

predict(mod, newdata, interval="predict")

confidence interval

> predict(mod, newdata, interval="confidence",level=.95)

fit lwr upr

1 81.03364 43.52379 118.5435

95% CI = (43.52, 118.54)

f). #prediction interval

> predict(mod, newdata, interval="predict",level=.95)

fit lwr upr

1 81.03364 14.19586 147.8714

95% PI=(14.20, 147.87)

g). No, there is not evidence this factor is significant. It should be dropped from the model.

4 0

2 years ago

What is one way to test whether an unknown solution is acidic or basic?

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Answer:

Bases change red litmus into blue and bases have a slippery and soapy texture. Therefore, we can conclude that out of the given options, one way to test whether an unknown solution is acidic or basic is to check whether the solution has a slippery feel

Step-by-step explanation:

3 0

2 years ago

With a height of 68 in, Nelson was the shortest president of a particular club in the past century. The club presidents of the

Ivahew [28]

Answer:

a. The positive difference between Nelson's height and the population mean is: $\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in$ .

b. The difference found in part (a) is 1.174 standard deviations from the mean (without taking into account if the height is above or below the mean).

c. Nelson's z-score: $\\ z = -1.1739 \approx -1.174$ (Nelson's height is below the population's mean 1.174 standard deviations units).

d. Nelson's height is usual since $\\ -2 < -1.174 < 2$ .

Step-by-step explanation:

The key concept to answer this question is the z-score. A z-score "tells us" the distance from the population's mean of a raw score in standard deviation units. A positive value for a z-score indicates that the raw score is above the population mean, whereas a negative value tells us that the raw score is below the population mean. The formula to obtain this z-score is as follows:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

$\\ z$ is the z-score.

$\\ \mu$ is the population mean.

$\\ \sigma$ is the population standard deviation.

From the question, we have that:

Nelson's height is 68 in. In this case, the raw score is 68 in $\\ x = 68$ in.
$\\ \mu = 70.7$ in.
$\\ \sigma = 2.3$ in.

With all this information, we are ready to answer the next questions:

a. What is the positive difference between Nelson's height and the mean?

The positive difference between Nelson's height and the population mean is (taking the absolute value for this difference):

$\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in$ .

That is, the positive difference is 2.7 in.

b. How many standard deviations is that [the difference found in part (a)]?

To find how many standard deviations is that, we need to divide that difference by the population standard deviation. That is:

$\\ \frac{2.7\;in}{2.3\;in} \approx 1.1739 \approx 1.174$

In words, the difference found in part (a) is 1.174 standard deviations from the mean. Notice that we are not taking into account here if the raw score, x, is below or above the mean.

c. Convert Nelson's height to a z score.

Using formula [1], we have

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{68\;in - 70.7\;in}{2.3\;in}$

$\\ z = \frac{-2.7\;in}{2.3\;in}$

$\\ z = -1.1739 \approx -1.174$

This z-score "tells us" that Nelson's height is 1.174 standard deviations below the population mean (notice the negative symbol in the above result), i.e., Nelson's height is below the mean for heights in the club presidents of the past century 1.174 standard deviations units.

d. If we consider "usual" heights to be those that convert to z scores between minus2 and 2, is Nelson's height usual or unusual?

Carefully looking at Nelson's height, we notice that it is between those z-scores, because:

$\\ -2 < z_{Nelson} < 2$

$\\ -2 < -1.174 < 2$

Then, Nelson's height is usual according to that statement.

7 0

3 years ago