Is -5.432.... rational or irrational

Earning for 3weeks from a part time job shown in the table . He is offered a job that will pay 7.25 per hour . Which job pays be

xz_007 [3.2K]

7.25 hour is better because multiply by eleven and its 79.75 which is higher than the on on the table

3 0

3 years ago

In a survey of randomly selected 3,900 family-owned businesses with revenues exceeding $1 million a year, it was found that 1,91

Maurinko [17]

Answer:

a) The 90% confidence interval to estimate the proportion of family-owned businesses without strategic business plans is (0.4768, 0.5032). This means that we are 90% sure that the true proportion of all family-owned businesses without strategic business plans is between these two values.

b) Wider

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

In a survey of randomly selected 3,900 family-owned businesses with revenues exceeding $1 million a year, it was found that 1,911 of them had no strategic business plan.

This means that $n = 3900, \pi = \frac{1911}{3900} = 0.49$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 - 1.645\sqrt{\frac{0.49*0.51}{3900}} = 0.4768$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 + 1.645\sqrt{\frac{0.49*0.51}{3900}} = 0.5032$

The 90% confidence interval to estimate the proportion of family-owned businesses without strategic business plans is (0.4768, 0.5032). This means that we are 90% sure that the true proportion of all family-owned businesses without strategic business plans is between these two values.

Question b:

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

The higher the confidence level, the higher the value of z, thus the higher the margin of error and the interval is wider. Thus, a 99% confidence interval is wider than a 90% confidence interval.

3 0

2 years ago

What is the percent of 16/25

Lelechka [254]

16/25 = 0.64 = 64%

Another way to do it is to multiply top and bottom by 4
16/25 = (16*4)/(25*4) = 64/100 = 64%

Either way, the answer is 64%

7 0

3 years ago

Read 2 more answers

Grayson is deciding between two truck rental companies. Company A charges an initial fee of $40 for the rental plus $2 per mile

Alexandra [31]

Answer:

A is 10 dollar cheaper than B

Step-by-step explanation:

A : 40 dollars plus 45*2=90

90+40= 130 dollars

B:100 dollars 45*1=45

100+45= 145 dollars

5 0

3 years ago

Read 2 more answers

Evaluate the difference quotient for the given function. Simplify your answer.

nasty-shy [4]

I suppose you mean

$f(x) = \dfrac{x+5}{x+1}$

Then

$f(3) = \dfrac{3+5}{3+1} = \dfrac84 = 2$

and the difference quotient is

$\dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5}{x+1}-2}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5-2(x+1)}{x+1}}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{-x+3}{(x+1)(x-3)} \\\\ \dfrac{f(x)-f(3)}{x-3} = \boxed{-\dfrac{x-3}{(x+1)(x-3)}}$

If it's the case that x ≠ 3, then (x - 3)/(x - 3) reduces to 1, and you would be left with

$\dfrac{f(x)-f(3)}{x-3}\bigg|_{x\neq3} = -\dfrac1{x+1}$

4 0

2 years ago