Sketch the graph of the equation y=3/8x+5

The producer of a certain bottling equipment claims that the variance of all its filled bottles is .027 or less. A sample of 30

o-na [289]

Answer:

$p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459$

b. between .025 and .05

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

$\bar X$ represent the sample mean

n = 30 sample size

s= 0.2 represent the sample deviation

$\sigma_o =\sqrt{0.027}=0.164$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic

$\alpha=$ significance level

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is less than 0.027, so the system of hypothesis are:

H0: $\sigma \leq 0.027$

H1: $\sigma >0.027$

In order to check the hypothesis we need to calculate the statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(30-1) [\frac{0.2}{0.164}]^2 =42.963$

What is the approximate p-value of the test?

The degrees of freedom are given by:

$df=n-1= 30-1=29$

For this case since we have a right tailed test the p value is given by:

$p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459$

And the best option would be:

b. between .025 and .05

6 0

3 years ago

Find the absolute maximum and minimum values of f(x, y) = x+y+ p 1 − x 2 − y 2 on the quarter disc {(x, y) | x ≥ 0, y ≥ 0, x2 +

Andreas93 [3]

Answer:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

Step-by-step explanation:

In order to find the absolute max and min, we need to analyse the region inside the quarter disc and the region at the limit of the disc:

Region inside the quarter disc:

There could be Minimums and Maximums, if:

∇f(x,y)=(0,0) (gradient)

we develop:

(1-2x, 1-2y)=(0,0)

x=1/2

y=1/2

Critic point P(1/2,1/2) is inside the quarter disc.

f(P)=1/2+1/2+p1-1/4-1/4=1/2+p1

f(0,0)=p1

We see that:

f(P)>f(0,0), then P(1/2,1/2) is a maximum relative

Region at the limit of the disc:

We use the Method of Lagrange Multipliers, when we need to find a max o min from a f(x,y) subject to a constraint g(x,y); g(x,y)=K (constant). In our case the constraint are the curves of the quarter disc:

g1(x, y)=x^2+y^2=1

g2(x, y)=x=0

g3(x, y)=y=0

We can obtain the critical points (maximums and minimums) subject to the constraint by solving the system of equations:

∇f(x,y)=λ∇g(x,y) ; (gradient)

g(x,y)=K

Analyse in g2:

x=0;

1-2y=0;

y=1/2

Q(0,1/2) critical point

f(Q)=1/4+p1

We do the same reflexion as for P. Q is a maximum relative

Analyse in g3:

y=0;

1-2x=0;

x=1/2

R(1/2,0) critical point

f(R)=1/4+p1

We do the same reflexion as for P. R is a maximum relative

Analyse in g1:

(1-2x, 1-2y)=λ(2x,2y)

x^2+y^2=1

Developing:

x=1/(2λ+2)

y=1/(2λ+2)

x^2+y^2=1

So:

(1/(2λ+2))^2+(1/(2λ+2))^2=1

$\lambda_{1}=\sqrt{1/2}*-1 =-0.29$

$\lambda_{2}=-\sqrt{1/2}*-1 =-1.71$

$\lambda_{2}$ give us (x,y) values negatives, outside the region, so we do not take it in account

For $\lambda_{1}$ : S(x,y)=(0.70, 070)

and

f(S)=0.70+0.70+p1-0.70^2-0.70^2=0.42+p1

We do the same reflexion as for P. S is a maximum relative

Points limits between g1, g2 y g3

we need also to analyse the points limits between g1, g2 y g3, that means U(0,0), V(1,0), W(0,1)

f(U)=p1

f(V)=p1

f(W)=p1

We can see that this 3 points are minimums relatives.

Conclusion:

We compare all the critical points P,Q,R,S,T,U,V,W an their respective values f(x,y). We find that:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

4 0

3 years ago

PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!! I NEED TO FINISH THESE QUESTIONS BEFORE MIDNIGHT TONIGHT.

Lina20 [59]

Yo sup??

we can solve this question by applying trigonometric ratios

cos59=CB/CD

CD=CB/cos59

=7.8

Hope this helps.

7 0

3 years ago

PLEASE HELP! i would really appreciate it thank you

LiRa [457]

Answer:

7

Step-by-step explanation:

5 0

3 years ago

The scatter plot below shows the relationship between the diameter (x), in mm, and mass f(x), in grams, of a berry: ordered pair

ivann1987 [24]

For this case the first thing we must observe is that the mass increases 0.4 grams when the diameter increases 1 millimeter.
Therefore, the slope of the line is given by:
m = 0.4
Thus, the function that best suits the table is given by:
f (x) = -4 + 0.4x
For example, for x = 20 we have:
f (20) = -4 + 0.4 (20)
f (20) = -4 + 8
f (20) = 4
The result, matches the table.
Answer:
The function that is best represented by the scatter plot is:
f (x) = -4 + 0.4x

6 0

3 years ago

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