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IrinaK [193]
2 years ago
6

Please help me with this question thank you !!

Mathematics
1 answer:
krok68 [10]2 years ago
8 0
The correct answer is B. 3 + 7 = 10 and (3 x 3) + 1 = 10
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The two other points are (-2,0)(2,0)
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If Sine (x) = two-fifths and tan(x) > 0, what is sin(2x)?
ElenaW [278]

B) \frac{4\sqrt{21}}{25}

Correct on Edge

4 0
3 years ago
Read 2 more answers
A manufacturer of detergent claims that the contents of boxes sold weigh on average at least 20 ounces. The distribution of weig
jenyasd209 [6]

Answer:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

Step-by-step explanation:

Information provided

\bar X=19.87 represent the sample mean

\sigma=0.4 represent the population deviation

n=25 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0.01 represent the significance level

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:  

Null hypothesis:\mu \geq 20  

Alternative hypothesis:\mu < 20  

The statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Replacing the info given we got:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

7 0
2 years ago
Consider the differential equation: xy′(x2+7)y=cos(x)+e3xy. Put the differential equation into the form: y′+p(x)y=g(x), determin
icang [17]

Answer:

Linear and non-homogeneous.

Step-by-step explanation:

We are given that

\frac{xy'}{(x^2+7)y}=cosx+\frac{e^{3x}}{y}

We have to convert into y'+P(x)y=g(x) and determine P(x) and g(x).

We have also find type of differential equation.

y'=\frac{(x^2+7)y}{x}(cosx+\frac{e^{3x}}{y}}

y'=\frac{(x^2+7)cosx}{x}y+\frac{(x^2+7)e^{3x}}{x}

y'-\frac{cosx(x^2+7)}{x}y=\frac{e^{3x}(x^2+7)}{x}

It is linear differential equation because  this equation is of the form

y'+P(x)y=g(x)

Compare it with first order first degree linear differential equation

y'+P(x)y=g(x)

P(x)=-\frac{cosx (x^2+7)}{x},g(x)=\frac{e^{3x}(x^2+7)}{x}

\frac{dy}{dx}=\frac{(x^2+7)(ycosx+e^{3x})}{x}

Homogeneous equation

\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}

Degree of f and g are same.

f(x,y)=(x^2+7)(ycosx+e^{3x}),g(x,y)=x

Degree of f and g are not same .

Therefore, it is non- homogeneous .

Linear and non-homogeneous.

3 0
3 years ago
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