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2 years ago

10

Debbie studies for 12.5 hours, John studies for 6 hours, and Ashley studies for 1.5 hours. what is the average number of hours t

he 3 students study?

1 answer:

ale4655 [162]2 years ago

6 0

you can answer it six point six six six six six infinity or you can answer it 6.7 depending upon how they want

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A random sample of 25 roundtrip flights between Philadelphia and Dallas has an average airfare of $393.50 with a sample standard

Sauron [17]

Answer:

$393.50+/-$19.72

= ( $373.78, $413.22)

Therefore, the 95% confidence interval (a,b) = ($373.78, $413.22)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = $393.50

Standard deviation r = $50.30

Number of samples n = 25

Confidence interval = 95%

z value(at 95% confidence) = 1.96

Substituting the values we have;

$393.50+/-1.96($50.30/√25)

$393.50+/-1.96($10.06)

$393.50+/-$19.7176

$393.50+/-$19.72

= ( $373.78, $413.22)

Therefore, the 95% confidence interval (a,b) = ($373.78, $413.22)

5 0

2 years ago

Can you define f(0, 0) = c for some c that extends f(x, y) to be continuous at (0, 0)? If so, for what value of c? If not, expla

Ahat [919]

(i) Yes. Simplify $f(x,y)$ .

$\displaystyle \frac{x^2 - x^2y^2 + y^2}{x^2 + y^2} = 1 - \frac{x^2y^2}{x^2 + y^2}$

Now compute the limit by converting to polar coordinates.

$\displaystyle \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = \lim_{r\to0} \frac{r^4 \cos^2(\theta) \sin^2(\theta)}{r^2} = 0$

This tells us

$\displaystyle \lim_{(x,y)\to(0,0)} f(x,y) = 1$

so we can define $f(0,0)=1$ to make the function continuous at the origin.

Alternatively, we have

$\dfrac{x^2y^2}{x^2+y^2} \le \dfrac{x^4 + 2x^2y^2 + y^4}{x^2 + y^2} = \dfrac{(x^2+y^2)^2}{x^2+y^2} = x^2 + y^2$

and

$\dfrac{x^2y^2}{x^2+y^2} \ge 0 \ge -x^2 - y^2$

Now,

$\displaystyle \lim_{(x,y)\to(0,0)} -(x^2+y^2) = 0$

$\displaystyle \lim_{(x,y)\to(0,0)} (x^2+y^2) = 0$

so by the squeeze theorem,

$\displaystyle 0 \le \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} \le 0 \implies \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = 0$

and $f(x,y)$ approaches 1 as we approach the origin.

(ii) No. Expand the fraction.

$\displaystyle \frac{x^2 + y^3}{xy} = \frac xy + \frac{y^2}x$

$f(0,y)$ and $f(x,0)$ are undefined, so there is no way to make $f(x,y)$ continuous at (0, 0).

(iii) No. Similarly,

$\dfrac{x^2 + y}y = \dfrac{x^2}y + 1$

is undefined when $y=0$ .

5 0

1 year ago

In ΔQRS, r = 89 inches, s = 32 inches and ∠Q=116°. Find ∠R, to the nearest degree.

allsm [11]

Answer is in the photo. I can't attach it here, but I uploaded it to a file hosting. link below! Good Luck!

tinyurl.com/wpazsebu

3 0

2 years ago

Glen received $2,250 loan from bank. After six months, he paid back $2,295 and closed the loan. Find the rate of interest

kumpel [21]

Answer: 6.824%

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8 0

2 years ago

What is the answer to this question

LiRa [457]

Answer:

Step-by-step explanation:

Tienes que multiplicar 10 ×7

6 0

2 years ago