7.489 divided by 0.09

How to find the circumference of a cirlcle

Otrada [13]

$\sf{Circumference = 2\times\pi\times~r}$

Where r is the radius of the circle.

7 0

3 years ago

find the equation of a cubic function whose graph passes through points (3,0) and (1,4) and is tangent to x-axis at the origin

Tomtit [17]

Answer:

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

Step-by-step explanation:

Preliminary Remark

If a cubic is tangent to the x axis at 0,0

Then the equation must be related to y = a*x^2(x - h)

(3,0)

If the cubic goes through the point (3,0), then the equation will become

0 = a*3^2(3 - h)

0 = 9a (3 - h)

0 = 27a - 9ah

from which h = 3

From the second point, we get

4 = ax^2(x - 3)

4 = a(1)^2(1 - 3)

4 = a(-2)

a = 4 / - 2

a = -2

Answer

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

3 0

2 years ago

If z varies inversely as w, and z= 3 when w=4, find z when w=2. Z=

vivado [14]

Answer:

6

Step-by-step explanation:

Inverse proportion.

z = k/w

z = 3, w = 4

Solve for k constant.

3 = k/4

(4)3 = (4)k/4

12=k

The value of k constant is 12.

z = k/w

Put w as 2 and k as 12 and solve for z.

z = 12/2

z = 6

The value of z when w equals 2 is 6.

8 0

3 years ago

Consider the following hypothesis test. H0: μ ≥ 55 Ha: μ < 55 A sample of 36 is used. Identify the p-value and state your con

Ket [755]

Answer:

Step-by-step explanation:

Given that:

$H_o: \mu \ge 55 \ \\ \\ H_1 : \mu < 55$

(a) For x = 54 and s = 5.3

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{54-55}{\dfrac{5.3}{\sqrt{36}} }$

$Z = \dfrac{-1}{\dfrac{5.3}{6} }$

Z = -1.132

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-1.132,35,1) = 0.1326

Decision: p-value is greater than significance level; do not reject $\mathbf{H_o}$

$Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \$ $\mu < 55$

b

For x = 53 and s = 4.6

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{53-55}{\dfrac{4.6}{\sqrt{36}} }$

$Z = \dfrac{-2}{\dfrac{4.6}{6} }$

Z = -2.6087

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-2.6087,35,1) =0.0066

Decision: p-value is < significance level; we reject the null hypothesis.

$Conclusion: \ There \ is \ sufficient \ evidence \ to \ conclude \ that$ $\mu < 55$

c)

For x = 56 and s = 5.0

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }$

Z = 1.2

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(1.2,35,1) = 0.88009

Decision: p-value is greater than significance level; do not reject $\mathbf{H_o}$

$Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \$ $\mu < 55$

6 0

3 years ago

Please help me with the process to find the answer for #15, #16, #17 and #18

timama [110]

Well I wanna do you want us to pick you guys up tomorrow at night to

8 0

3 years ago