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Artemon [7]
3 years ago
6

Write an equation of the circle with center (6, 2) and radius 4.

Mathematics
2 answers:
luda_lava [24]3 years ago
8 0

Answer:

(x-6)^2+(y-2)^2=16.

Step-by-step explanation:

(x-h)^2+(y-k)^2=r^2 is the equation for a circle with center (h,k) and radius r.

You are given center (6,2) and radius 4.

So we will replace h with 6 and k with 2 and r with 4.

This gives us:

(x-6)^2+(y-2)^2=4^2

Simplify:

(x-6)^2+(y-2)^2=16.

UNO [17]3 years ago
4 0

For this case we have that by definition, the equation of a circle is given by:

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

Where:

(h, k):It is the center of the circle

r: It is the radius of the circle

According to the data we have to:

(h, k) :( 6.2)\\r = 4

Substituting:

(x-6) ^ 2 + (y-2) ^ 2 = 4 ^ 2\\(x-6) ^ 2 + (y-2) ^ 2 = 16

ANswer:

(x-6) ^ 2 + (y-2) ^ 2 = 16

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A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random
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Answer:

a. Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 >\mu_2

b. t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282

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So with the p value obtained 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, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).  

Step-by-step explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 >\mu_2

Or equivalently:

Null hypothesis: \mu_1 - \mu_2 \leq 0

Alternative hypothesis: \mu_1 -\mu_2>0

Our notation on this case :

n_1 =78 represent the sample size for group 1

n_2 =21 represent the sample size for group 2

\bar X_1 =92.88 represent the sample mean for the group 1

\bar X_2 =86.90 represent the sample mean for the group 2

s_1=15.34 represent the sample standard deviation for group 1

s_2=8.99 represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom. If we replace the values given we have:

t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282

Now we can calculate the degrees of freedom given by:

df=78+21-2=97

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

p_v =P(t_{97}>2.287) =0.0122

So with the p value obtained 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, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).  

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3 years ago
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