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3 years ago

4x + 3 = x + 2(x + 7) x=

Mathematics

1 answer:

stiv31 [10]3 years ago

4 0

Answer:

X= -11/4 + -1/4 √145 or X= -11/4 + 1/4 √145

Step-by-step explanation:

Hope this helped<3

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dlinn [17]

$\\\\\ \textbf{a)}\\\\~~~\displaystyle \int (6x- \sin 3x) ~ dx\\\\=6\displaystyle \int x ~ dx - \displaystyle \int \sin 3x ~ dx\\\\=6 \cdot \dfrac{x^2}2 - \dfrac 13 (- \cos 3x) +C~~~~~~~~~~~;\left[\displaystyle \int x^n~ dx = \dfrac{x^{n+1}}{n+1}+C,~~~n \neq -1\right]\\\\ =3x^2 +\dfrac{\cos 3x}3 +C~~~~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \sin (mx) ~dx = -\dfrac 1m ~ (\cos mx)+C \right]\\$

$\textbf{b)}\\\\~~~~\displaystyle \int(3e^{-2x} +\cos (0.5 x)) dx\\\\=3\displaystyle \int e^{-2x} ~dx+ \displaystyle \int \cos(0.5 x) ~dx\\\\\\=-\dfrac 32 e^{-2x} + \dfrac 1{0.5} \sin (0.5 x) +C~~~~~~~~~~~~~~;\left[\displaystyle \int e^{mx}~dx = \dfrac 1m e^{mx} +C \right]\\\\\\=-\dfrac 32 e^{-2x} + 2 \sin(0.5 x) +C~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \cos(mx)~ dx = \dfrac 1m \sin(mx) +C\right]\\\\\\=-1.5e^{-2x} +2\sin(0.5x) +C$

$\textbf{a)}\\\\y = \displaystyle \int \cos(x+5) ~ dx\\\\\text{Let,}\\\\~~~~~~~u = x+5\\\\\implies \dfrac{du}{dx} = 1+0~~~~~~;[\text{Differentiate both sides.}]\\\\\implies \dfrac{du}{dx} = 1\\\\\implies du = dx\\\\\text{Now,}\\\\y= \displaystyle \int \cos u ~ du\\\\~~~= \sin u +C\\\\~~~=\sin(x+5) + C$

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$\textbf{a)}\\\\y = \displaystyle \int xe^{3x} dx\\\\\text{We know that,}\\\\ \displaystyle \int (uv) ~dx = u \displaystyle \int v ~ dx - \displaystyle \int \left[ \dfrac{du}{dx} \displaystyle \int ~ v ~ dx \right]~ dx\\\\\text{Let}, u =x~ \text{and}~ v=e^{3x} .\\\\y= \displaystyle \int xe^{3x} ~dx\\\\\\~~= x\displaystyle \int e^{3x} ~ dx - \displaystyle \int \left[\dfrac{d}{dx}(x) \displaystyle \int e^{3x}~ dx \right]~ dx\\\\\\$

$=x\displaystyle \int e^{3x}~ dx - \displaystyle \dfrac 13 \int \left(e^{3x} \right)~ dx\\\\\\=\dfrac{xe^{3x}}3 - \dfrac 13 \cdot \dfrac{ e^{3x}}3+C\\\\\\= \dfrac{xe^{3x}}{3}- \dfrac{e^{3x}}{9}+C\\\\\\=\dfrac{3xe^{3x}}{9}- \dfrac{e^{3x}}9 + C\\\\\\= \dfrac 19e^{3x}(3x-1)+C$

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8 0

2 years ago

The bad debt ratio for a financial institution is defined to the dollar value of loans defaulted divided by the total dollar val

vlabodo [156]

Answer:

Step-by-step explanation:

From the information given:

Consider X to be the random variable denoting the bad debt ratios for Ohio Bank.

Then, $X \sim N ( \mu, \sigma ^2)$

Thus the null hypothesis and the alternative can be computed as:

Null hypothesis:

$H_o : \mu \leq3.5\%$

Alternative hypothesis

$H_1 : \mu > 3.5\%$

The type I and type II error is as follows:

Type I:

The mean bad debt ratio is > 3.5% when it is not

Type II:

The mean bad debt ratio is ≤ 3.5% when it is not.

The test statistics can be calculated by using the formula:

$t = \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt{n}}}$

where;

sample size n = 7

mean = 6+8+5+9+7+5+8 = 48

sample mean $\overline x =\dfrac{48}{7}$

$\overline x$ = 6.86

sample standard deviation is :

$s = \sqrt{\dfrac{\sum( x -\overline x)^2}{n-1}}$

$s = \sqrt{\dfrac{( 6 -6.86)^2+( 8-6.86)^2+ ( 5 -6.86)^2 + ...+( 7 -6.86)^2+ ( 5 -6.86)^2+( 8 -6.86)^2 }{7-1}}$

s = 1.573

population mean $\mu = 3.5$

Therefore, the test statistics is :

$t = \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt{n}}}$

$t = \dfrac{6.86- 3.5}{\dfrac{1.573}{\sqrt{7}}}$

$t = \dfrac{3.36}{\dfrac{1.573}{2.65}}$

$t = \dfrac{3.36\times {2.65}}{{1.573}}$

t = 5.660

At significance level of 0.01

$t_{0.01} = 3.707$

P - value = P(T > 5.66)

P - value = 1 - (T < 5.66)

P - value = 1 - 0.9993

P-value = 0.0007

Therefore, since $t_{0.01} < t$ , we reject the null hypothesis and conclude that the claim that the mean bad debt ratio for Ohio banks is higher than the mean for all financial institutions is true.