How do i solve -5w^4·6w

Find the y intercept of the line on the graph

saul85 [17]

It’s the vertical line on the graph

4 0

1 year ago

The width of a rectangle is 75% the length of the rectangle. the perimeter of the rectangle is 42 inches. a circle is drawn that

matrenka [14]

Answer:

x = 56.25

Step-by-step explanation:

The sides of the rectangle are in the ratio 3:4, so those together with the diagonal of the rectangle make a 3-4-5 right triangle. The sum of these adjacent-side ratio units is 3+4 = 7, so the perimeter in ratio units is 2×7 = 14. The actual perimeter is 42 inches, so each ratio unit must stand for ...

(42 in)/(14 units) = 3 in/unit

The diagonal of the rectangle is the diameter of the circle. Its length is 5 units, or ...

(5 units)×(3 in/unit) = 15 inches.

Then the radius is half the diameter, or 7.5 inches.

The area of a circle is given by ...

A = πr² = π(7.5 in)² = 56.25π in²

The value of x is 56.25.

4 0

3 years ago

Math work geometry help please

zaharov [31]

Answer:

(C)

Step-by-step explanation:

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7 0

2 years ago

Find the volume of the prism.

Gemiola [76]

Answer:

Step-by-step explanation:

The volume of the triangular prism, this is an isosceles triangular prism is

V=1/2 * a*b*h

V=1/2 x10x8x14=560 cm^3

4 0

2 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