A 10-foot board rests against a wall. The angle that the board makes with the

Q2 find the degree of the polynomial x4(4x+1)

jasenka [17]

Step-by-step explanation:

x⁴(4x+1)

4x⁵+x⁴

degree is 5

3 0

3 years ago

Find the missing side lengths. Can someone please help???

adoni [48]

Answer:

x = 16

y = 13.8

Step-by-step explanation:

6 0

3 years ago

within a kiambu county,students were randomly assigned to one of two mathematics teachers.Mrs.Elite and Mrs. Bright. After the a

DerKrebs [107]

Using the t-distribution, as we have the standard deviation for the samples, it is found that since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, we test if they are equally effective teachers, that is, the subtraction of their means is 0, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, we test if they are not equally effective teachers, that is, the subtraction of their means is not 0, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

<h3>What is the distribution of the differences?</h3>

For Mrs. Elite, we have that:

$\mu_1 = 78, \sigma_1 = 10, n_1 = 30, s_1 = \frac{10}{\sqrt{30}} = 1.82574$

For Mrs. Bright, we have that:

$\mu_2 = 85, \sigma_2 = 15, n_2 = 25, s_2 = \frac{15}{\sqrt{25}} = 3$

For the distribution of differences, we have that:

$\overline{x} = \mu_1 - \mu_2 = 78 - 85 = -7$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{1.82574^2 + 3^2} = 3.5119$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{-7 - 0}{3.5119}$

$t = -1.99$

Considering a two-tailed test, as we are testing if the mean is different of a value, with 30 + 25 - 2 = 53 df and a significance level of 0.1, the critical value is of $|t^{\ast}| = 1.6741$ .

Since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

To learn more about the t-distribution, you can take a look at brainly.com/question/13873630

3 0

3 years ago

Need help please its Calculus. Ill give the 5 stars as well.

algol13

Answer:

$\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1$

General Formulas and Concepts:

Pre-Algebra

Order of Operations
Equality Properties

Algebra I

Functions
Function Notation
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Algebra II

Natural logarithms ln and Euler's number e

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Slope Fields

Separation of Variables

Solving Differentials

Integrals

Antiderivatives

Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

U-Substitution

Logarithmic Integration: $\displaystyle \int {\frac{1}{u}} \, dx = ln|u| + C$

Step-by-step explanation:

*Note:

When solving differential equations in slope fields, disregard the integration constant C for variable y.

Step 1: Define

$\displaystyle \frac{dy}{dx} = x^2(y - 1)$

$\displaystyle f(0) = 3$

Step 2: Rewrite

Separation of Variables. Get differential equation to a form where we can integrate both sides and rewrite Leibniz Notation.

[Separation of Variables] Rewrite Leibniz Notation: $\displaystyle dy = x^2(y - 1) \ dx$
[Separation of Variables] Isolate y's together: $\displaystyle \frac{1}{y - 1} \ dy = x^2 \ dx$

Step 3: Find General Solution Pt. 1

[Differential] Integrate both sides: $\displaystyle \int {\frac{1}{y - 1}} \, dy = \int {x^2} \, dx$
[dx Integral] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle \int {\frac{1}{y - 1}} \, dy = \frac{x^3}{3} + C$

Step 4: Find General Solution Pt. 2

Identify variables for u-substitution for dy.

Set: $\displaystyle u = y - 1$
Differentiate [Basic Power Rule]: $\displaystyle du = dy$

Step 5: Find General Solution Pt. 3

[dy Integral] U-Substitution: $\displaystyle \int {\frac{1}{u}} \, du = \frac{x^3}{3} + C$
[dy Integral] Integrate [Logarithmic Integration]: $\displaystyle ln|u| = \frac{x^3}{3} + C$
[Equality Property] e both sides: $\displaystyle e^\bigg{ln|u|} = e^\bigg{\frac{x^3}{3} + C}$
Simplify: $\displaystyle |u| = Ce^\bigg{\frac{x^3}{3}}$
Rewrite: $\displaystyle u = \pm Ce^\bigg{\frac{x^3}{3}}$
Back-Substitute: $\displaystyle y - 1 = \pm Ce^\bigg{\frac{x^3}{3}}$
[Equality Property] Isolate y: $\displaystyle y = \pm Ce^\bigg{\frac{x^3}{3}} + 1$

General Form: $\displaystyle y = \pm Ce^\bigg{\frac{x^3}{3}} + 1$

Step 6: Find Particular Solution

Substitute in function values [General Form]: $\displaystyle 3 = \pm Ce^\bigg{\frac{0^3}{3}} + 1$
Simplify: $\displaystyle 3 = \pm C + 1$
[Equality Property] Isolate C: $\displaystyle 2 = \pm C$
Rewrite: $\displaystyle C = 2$
Substitute in C [General Form]: $\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1$