Marvella writes the addition problem 5 + 6. Is this sum even or odd

In a Harris poll of 514 human resource professionals, 90% said that the appearance of a job applicant is most important for a go

jolli1 [7]

Answer:

( a ) 463 " human resource professionals " believe that the appearance of a job applicant is most important for a good first impression

( b ) Percent wise the confidence interval should be from about 86.6% to 93.4%

Step-by-step explanation:

I believe these first two parts here is sufficient enough to get you started!

" In a Harris poll of 514 human resource professionals, 90% said that the appearance of a job applicant is most important for a good first impression. Part A: Among the 514 human resource professionals who were surveyed, how many of them said that the appearance of a job applicant is most important for a good first impression?

Part B: Construct a 99% confidence interval estimate of the proportion of all human resource professionals believing that the appearance of a job applicant is most important for a good first impression. "

_____

( a ) We are given that n = 514, and p = 90%. The number of human resource professionals that said that the appearance of a job applicant is most important for a good first impression, should be the following -

514 x $\frac{90}{100}$

= $\frac{46260}{100}$

= 462.6

As the count of people can't be expressed as a fraction, the solution should be about 463 " human resource professionals. "

_____

( b ) Here the confidence level is 0.99 percent. Knowing that -

1 - ∝ = 0.99,

∝ = 1 - 0.99... = 0.01

Therefore, a 99% confidence interval estimate of the proportion of all human resource professionals believing that the appearance of a job applicant is most important for a good first impression should be calculated as directed in the attachment. By that the interval ranges as such -

0.8659 < p < 0.9341

And percent wise the confidence interval should be from about 86.6% to 93.4%

8 0

3 years ago

Divide the polynomials. Your answer should be in the form p(x) + k/x where p is a polynomials and k is an integer. 4x^4-x^2+3/x

kolezko [41]

Answer:

Step-by-step explanation:

4 $x^{4}$ -x²+3 / x =

So, we need to think about this problem as either what we have to multiply x by to get the equation, and do long division

= 4x³- x + $\frac{3}{x}$

6 0

3 years ago

Same question for this. Actually need help... ;-;

Bess [88]

372 is the answer.
(9*6*3)+(10*7*3)=372

7 0

3 years ago

On a final exam, each multiple-choice question is worth 4 points and each word problem is worth 8 points. Lorenzo needs

Elena-2011 [213]

Answer:

4x + 8y = 50 or 4x + 8y > 50

Step-by-step explanation:

To find which equation show the exact answer, Becasue he need at least 50 poitns to have a B, so:

4x + 8y must larger or equal to 50

In these choice, some I can't figure out what it means but the correct answer for this is:

4x + 8y = 50 or 4x + 8y > 50

Hope this help you :3

7 0

4 years ago

Read 2 more answers

Evaluate Dx / ^ 9-8x - x2^

Solnce55 [7]

It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for $\sqrt x$ .

In that case, you want to find the antiderivative,

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}$

Complete the square in the denominator:

$9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2$

Now substitute $x+4=5\sin y$ , so that $\mathrm dx=5\cos y\,\mathrm dy$ . Then

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy$

which simplifies to

$\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy$

Now, recall that $\sqrt{x^2}=|x|$ . But we want the substitution we made to be reversible, so that

$x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)$

which implies that $-\dfrac\pi2\le y\le\dfrac\pi2$ . (This is the range of the inverse sine function.)

Under these conditions, we have $\cos y\ge0$ , which lets us reduce $\sqrt{\cos^2y}=|\cos y|=\cos y$ . Finally,

$\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C$

and back-substituting to get this in terms of $x$ yields

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C$

4 0

3 years ago