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3 years ago

Either my brain is straight up goneskiis or I need assistance.

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

4 0

The answer is C. 89 in^2

This is found by (1/2( $\pi r^{2}$ ))+70

r in this equation stands for radius which is half the diameter. Sice the diameter equals 7, the radius equals 3.5

You have to solve this by putting it into a calculator since you can't really solve for an equation with pi in it otherwise.

Hope this helps!

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Suppose that the hourly wages of fast food workers are normally distributed with an unknown mean and standard deviation. The wag

Nutka1998 [239]

Answer: 1.303639

Step-by-step explanation:

The t-score for a level of confidence $(1-\alpha)$ is given by :_

$t_{(df,\alpha/2)}$ , where df is the degree of freedom and $\alpha$ is the significance level.

Given : Level of significance : $1-\alpha:0.80$

Then , significance level : $\alpha: 1-0.80=0.20$

Sample size : $n=40$

Then , the degree of freedom for t-distribution: $df=n-1=40-1=39$

Using the normal t-distribution table, we have

$t_{(df,\alpha/2)}=t_{39,0.10}=1.303639$

Thus, the t-score should be used to find the 80% confidence interval for the population mean =1.303639

8 0

3 years ago

Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x

kykrilka [37]

Parameterize $S$ by

$\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take a normal vector to $S$ ,

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k$

which has norm

$\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v$

Then the integral of $f(x,y,z)=x^2+y^2$ over $S$ is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$

$=\displaystyle36^2\int_0^{\pi/2}\int_0^{2\pi}\sin^3v\,\mathrm du\,\mathrm dv=\boxed{1728\pi}$

6 0

4 years ago

True or false: It is possible to spend your limit on a credit card.

LenaWriter [7]

It is posable to spend ur limit on ur credit card

4 0

3 years ago

Read 2 more answers

If tan tetha =8/15,find the value of sin tetha+cos tetha all divided by cos tetha (1-cos tetha)

Dahasolnce [82]

Answer: 195.5 or 12 7/32

Step-by-step explanation:

There is no letter tetha in the table so I use α instead. However it is not sence to final result.

The expression is:

(sinα+cosα)/(cosα*(1-cosα))

Lets divide the nominator and denominator by cosα

(sinα/cosα+cosα/cosα)/(cosα*(1-cosα)/cosα)= (tanα+1)/(1-cosα)=

=(8/15+1)/(1-cosα)= 23/(15*(1-cosα)) (1)

As known cos²α=1-sin²α (divide by cos²α both sides of equation)

cos²a/cos²α=1/cos²α-sin²α/cos²α

1=1/cos²α-tg²α

1/cos²α=1+tg²α

cos²α=1/(1+tg²α)

cosα=sqrt(1/(1+tg²α))= +-sqrt(1/(1+64/225))=+-sqrt(225/(225+64))=

=+-sqrt(225/289)=+-15/17 (2)

Substitute in (1) cosα by (2):

1st use cosα=15/17

1) 23/(15*(1-cosα)) =23/(15*(1-15/17))= 23*17/2=195.5

2-nd use cosα=-15/17

2)23/(15*(1-cosα)) =23/(15*(1+15/17))= 23*17/32=12 7/32

7 0

3 years ago

Is a -2/3 a integer?Is it also a rational number?

Elena-2011 [213]

No it’s not a rational number

7 0

3 years ago