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3 years ago

AdhkjAhkmhdj help me please

Mathematics

1 answer:

kirza4 [7]3 years ago

7 0

4x+2x=84
3x+5y=126
5x+4y=?
--------------------
cerco la x nella prima equazione:
4x+2y=84
x=(84-2y)/4
x=(42-y)/2
-------------------------------------------------------------------------
cerco la y nella seconda equazione, sostituendo la x:
3x+5y=126
3[(42-y)/2]+5y=126
y= -18
------------------------------------------------------------------------
cerco la x della nella equazione, sostituendo la y:
4x+2y=84
4x+2(-18)=84
x=30
---------------------------------------------------------------------------------------------------
cerco i totale nella terza equazione, sostituendo i valori della x e della y:
5x+4y= ?
5*30 + 4*(-18)= 150-72 = 78

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Suppose that at a state college, a random sample of 41 students is drawn, and each of the 41 students in the sample is asked to

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Answer:

The confidence interval for 90% confidence would be narrower than the 95% confidence

Step-by-step explanation:

From the question we are told that

The sample size is n = 41

For a 95% confidence the level of significance is $\alpha = [100 - 95]\% = 0.05$ and

the critical value of $\frac{\alpha }{2}$ is $Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} }= 1.96$

For a 90% confidence the level of significance is $\alpha = [100 - 90]\% = 0.10$ and

the critical value of $\frac{\alpha }{2}$ is $Z_{\frac{\alpha }{2} } =Z_{\frac{0.10 }{2} }= 1.645$

So we see with decreasing confidence level the critical value decrease

Now the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }$

given that other values are constant and only $Z_{\frac{\alpha }{2} }$ is varying we have that

$E\ \ \alpha \ \ Z_{\frac{\alpha }{2} }$

Hence for reducing confidence level the margin of error will be reducing

The confidence interval is mathematically represented as

$\= x - E < \mu < \= x + E$

Now looking at the above formula and information that we have deduced so far we can infer that as the confidence level reduces , the critical value reduces, the margin of error reduces and the confidence interval becomes narrower

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2 years ago

What is 3a+22.5h*37q? im lost plz help

erik [133]

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2 years ago

Maria received 5/7 of the 42 votes cast for class president. how many votes did she receive?

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3 years ago

The supreme shipping company can load trucks with both rectangular and cylindrical containers. A rectangular container has a vol

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the number of rectangular containers to maximze their income is 18 while the number of cylindrical containers to maximize income is 12.

The maximum income is $1740

Let x be the number of rectangular containers and y the number of cylindrical containers.

Since a rectangular container has a volume of 100 ft ³and weighs 200 pounds and a cylindrical container has a volume of 200 ft.³ and weighs 100 pounds, and each truck has room for at most 4200 ft.³ of containers and can carry a maximum of 4800 pounds,

We have that the maximum volume of the truck V = 100x + 200y.

Also, the maximum weight of the truck is W = 200x + 100y

Since V = 4200 ft.³ and W = 4800 pounds,

100x + 200y = 4200 and 200x + 100y = 4800

x + 2y = 42 (1) and 2x + y = 48 (2)

Multiplying (1) by 2 and (2) by 1, we have

2x + 4y = 84 (3) and 2x + y = 48 (4)

Subtracting (4) from (3), we have

2x + 4y = 84

2x + y = 48

3y = 36

y = 36/3

y = 12

Substituting y into (1), we have

x + 2y = 42

x + 2(12) = 42

x + 24 = 42

x = 42 - 24

x = 18

Also, since the shipping company charges $60 for a rectangular container and $55 for a cylindrical container, the income, P = 60x + 55y

Since x = 18 and y = 12, the maximum income is P = 60x + 55y

= 60(18) + 55(12)

= 1080 + 660

= $1740

So, the number of rectangular containers to maximze their income is 18 while the number of cylindrical containers to maximize income is 12.

The maximum income is $1740

Learn more about maximum income here:

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