Solve for system of equations -x+2y=8, x-2y=-8

What is the simplified from of square root of 6x^16?

melomori [17]

The answer for the question in the picture is in option B.

The square root of 64 is 8 (because 8*8=64), and the square root of x^16 is x^8 (because x^8 times x^8 is x^16). Put these together to get the answer: 8x^8

7 0

3 years ago

49 43.2 cm is equal to: A 4.32 mm C 0.432 mm B 432 mm D 4320 mm

LekaFEV [45]

43.2cm is equal to 432mm.

6 0

4 years ago

Read 2 more answers

Find the jacobian of the transformation. x = u2 + uv, y = uv2

atroni [7]

$\mathbf J=\begin{bmatrix}\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{bmatrix}=\begin{bmatrix}2u+v&u\\v^2&2uv\end{bmatrix}$

The Jacobian has determinant

$\det\mathbf J=\begin{vmatrix}2u+v&u\\v^2&2uv\end{vmatrix}=(2u+v)2uv-uv^2=4u^2v+uv^2=uv(4u+v)$

5 0

4 years ago

The diagram below gives the dimensions of a swimming pool. 18 ft , 20ft, 36ft (c) What is the total area of the swimming pool? E

WARRIOR [948]

The picture in the attached figure

Part 1) What is the total area of the swimming pool?

we know that
area of the swimming pool=area rectangle-area semi circle

area rectangle=20*36-----> 720 ft²

area semicircle=pi*r²/2
r=18/2----> 9 ft
area semicircle=pi*9²/2----> 127.17 ft²

area of the swimming pool=720 ft²-127.17 ft²----> 592.83 ft²

the answer Part 1) is
The area of the swimming pool is 592.83 ft²

Part 2) What is the perimeter of the swimming pool?

perimeter of the swimming pool=perimeter of rectangle-18 ft+perimeter semi circle
perimeter of rectangle=2*[20+36]---> 112 ft

perimeter semi circle=2*pi*r/2----> pi*r
r=9 ft
perimeter semi circle=pi*9----> 28.26 ft
so
perimeter of the swimming pool=112 ft-18 ft+28.26 ft----> 122.26 ft

the answer Part 2) is
122.26 ft

6 0

3 years ago

Test the claim that the mean GPA of night students is larger than 2 at the .025 significance level. The null and alternative hyp

exis [7]

Answer:

$H_0: \, \mu = 2$ .

$H_1:\, \mu > 2$ .

Test statistics: $z \approx 2.582$ .

Critical value: $z_{1 - 0.025} \approx 1.960$ .

Conclusion: reject the null hypothesis.

Step-by-step explanation:

The claim is that the mean $\mu$ is greater than $2$ . This claim should be reflected in the alternative hypothesis:

$H_1:\, \mu > 2$ .

The corresponding null hypothesis would be:

$H_0:\, \mu = 2$ .

In this setup, the null hypothesis $H_0:\, \mu = 2$ suggests that $\mu_0 = 2$ should be the true population mean of GPA.

However, the alternative hypothesis $H_1:\, \mu > 2$ does not agree; this hypothesis suggests that the real population mean should be greater than $\mu_0= 2$ .

One way to test this pair of hypotheses is to sample the population. Assume that the population mean is indeed $\mu_0 = 2$ (i.e., the null hypothesis is true.) How likely would the sample (sample mean $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ ) be observed in this hypothetical population?

Let $\sigma$ denote the population standard deviation.

Given the large sample size $n = 60$ , the population standard deviation should be approximately equal to that of the sample:

$\sigma \approx s = 0.06$ .

Also because of the large sample size, the central limit theorem implies that $Z= \displaystyle \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}}$ should be close to a standard normal random variable. Use a $Z$ -test.

Given the observation of $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ :

$\begin{aligned}z_\text{observed}&= \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} \\ &\approx \frac{\overline{X} - \mu_0}{s / \sqrt{n}} = \frac{2.02 - 2}{0.06 / \sqrt{60}} \approx 2.582\end{aligned}$ .

Because the alternative hypothesis suggests that the population mean is greater than $\mu_0 = 2$ , the null hypothesis should be rejected only if the sample mean is too big- not too small. Apply a one-sided right-tailed $z$ -test. The question requested a significant level of $0.025$ . Therefore, the critical value $z_{1 - 0.025}$ should ensure that $P( Z > z_{1 - 0.025}) = 0.025$ .

Look up an inverse $Z$ table. The $z_{1 - 0.025}$ that meets this requirement is $z_{1 - 0.025} \approx 1.960$ .

The $z$ -value observed from the sample is $z_\text{observed}\approx 2.582$ , which is greater than the critical value. In other words, the deviation of the sample from the mean in the null hypothesis is sufficient large, such that the null hypothesis needs to be rejected at this $0.025$ confidence level in favor of the alternative hypothesis.

3 0

3 years ago