Which equation represents this statement?

Studentka2010 [4]

The answer is 2 i hope this helps

7 0

3 years ago

Given that the diameter of Circle A is 6 cm , and the radius of Circle B is 18 cm , what can be concluded about the two circles?

ELEN [110]

Answer:

The radius of circle B is 6 times greater than the radius of circle A

The area of circle B is 36 times greater than the area of circle A

Step-by-step explanation:

we have

<em>Circle A</em>

$D=6\ cm$

The radius of circle A is

$r=6/2=3\ cm$ -----> the radius is half the diameter

<em>Circle B</em>

$r=18\ cm$

Compare the radius of both circles

$3\ cm< 18\ cm$

$18=6(3)$

The radius of circle B is six times greater than the radius of circle A

Remember that , if two figures are similar, then the ratio of its areas is equal to the scale factor squared

All circles are similar

In this problem the scale factor is 6

so

$6^{2}=36$

therefore

The area of circle B is 36 times greater than the area of circle A

8 0

3 years ago

Dont skip help me plzzz will mark brainliest

Yakvenalex [24]

Answer:

(6,-6)

Step-by-step explanation:

well if you count over to the right 6 and down 6 that points would be

(6,-6)

3 0

3 years ago

Read 2 more answers

Simplify \sqrt{240}.......

Lesechka [4]

Answer:

$4\sqrt{15}$

7 0

2 years ago

Tensile-strength tests were carried out on two different grades of wire rod (Fluidized Bed Patenting of Wire Rods, Wire J., June

Shtirlitz [24]

Answer:

$t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569$

$p_v =P(t_{256}>28.569) \approx 0$

So with the p value obtained and using the significance level given $\alpha=0.1$ 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 true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.

Step-by-step explanation:

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

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2 +10$

Alternative hypothesis: $\mu_1 >\mu_2 +10$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 \leq 10$

Alternative hypothesis: $\mu_1 -\mu_2>10$

Our notation on this case :

$n_1 =129$ represent the sample size for group AISI 1078

$n_2 =129$ represent the sample size for group AISI 1064

$\bar X_1 =123.6$ represent the sample mean for the group AISI 1078

$\bar X_2 =107.6$ represent the sample mean for the group AISI 1064

$s_1=2.0$ represent the sample standard deviation for group 1 AISI 1078

$s_2=1.3$ represent the sample standard deviation for group AISI 1064

And now we can calculate the statistic:

$t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569$

Now we can calculate the degrees of freedom given by:

$df=129+129-2=256$

And now we can calculate the p value using the altenative hypothesis:

$p_v =P(t_{256}>28.569) \approx 0$

So with the p value obtained and using the significance level given $\alpha=0.1$ 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 true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.

7 0

3 years ago