A^4/2a^2-4a^4×8a^3+1/2a+4a^2

In order to circumscribe a circle on a triangle, which line must you construct?

SIZIF [17.4K]

<span>D) perpendicular bisector <em>I believe.
</em></span>

6 0

4 years ago

Serenity picked 75% of the apples of a tree in her backyard. If she picked 30 apples, how many apples were on the tree to begin

Stells [14]

Given:

Serenity picked 75% of the apples of a tree in her backyard.

She picked 30 apples.

To find:

The number of apples on the tree in the beginning.

Solution:

Let x be the number of apples on the tree in the beginning.

Serenity picked 75% of the apples of a tree in her backyard.

Number of apples picked = 75% of x

= $\dfrac{75}{100}x$

= $\dfrac{3}{4}x$

She picked 30 apples.

$\dfrac{3}{4}x=30$

Multiply both sides by 4.

$3x=4\times 30$

$3x=120$

Divide both sides by 3.

$x=\dfrac{120}{3}$

$x=40$

Therefore, the number of apples on the tree in the beginning is 40.

3 0

3 years ago

How can i write 1 as a power in three different ways?

Svetradugi [14.3K]

2^1 1^2
3^1 or 1^3
4^1 1^4
I'm not sure which.

5 0

4 years ago

In a study of the effect of cooling rate on the hardness of welded joints, 50 welds cooled at a rate of 10°C/s had an average Ro

Studentka2010 [4]

Answer:

a) The 95% confidence interval would be given by $-1.887 \leq \mu_1 -\mu_2 \leq 2.687$

b) No contradict the result obtained since the confidence interval contains the value 0, so we don't have enough evidence to conclude that we have a significant effect in the colling rate and the hardness.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =91.1$ represent the sample mean 1

$\bar X_2 =90.7$ represent the sample mean 2

n1=50 represent the sample 1 size

n2=40 represent the sample 2 size

$s_1 =6.55$ sample standard deviation for sample 1

$s_2 =4.32$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Part a

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =91.1-90.7=0.4$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_1 +n_2 -1=50+40-2=88$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,88)".And we see that $t_{\alpha/2}=1.987$

The standard error is given by the following formula:

$SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{6.55^2}{50}+\frac{4.32^2}{40}}=1.325$

Now we have everything in order to replace into formula (1):

$0.4-1.987\sqrt{\frac{6.55^2}{50}+\frac{4.32^2}{40}}=-1.887$

$0.4+1.987\sqrt{\frac{6.55^2}{50}+\frac{4.32^2}{40}}=2.687$

So on this case the 95% confidence interval would be given by $-1.887 \leq \mu_1 -\mu_2 \leq 2.687$

Part b

No contradict the result obtained since the confidence interval contains the value 0, so we don't have enough evidence to conclude that we have a significant effect in the colling rate and the hardness.

3 0

3 years ago

What number divided by 9 equals 12

True [87]

108 divided by 9 = 12.
you can just get the answer by multiplying 12 and 9 and you end getting 108. you can that answer by dividing after you multiply

6 0

4 years ago

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