Aryonna rolls a standard number cube once. How many possible outcomes be there?

mote1985 [20]

$\displaystyle\lim_{x\to1}\frac{\ln x}{\sin7\pi x}$

Note that both the numerator and denominator approach 0 as $x\to1$ , so we can try using L'Hopital's rule.

$\displaystyle\lim_{x\to1}\frac{\lnx }{\sin7\pi x}=\lim_{x\to1}\frac{\frac1x}{7\pi\cos7\pi x}=\lim_{x\to1}\frac1{7\pi x\cos7\pi x}$

The denominator is nonzero at $x=1$ , so the limit is equivalent to

$\displaystyle\frac1{\lim\limits_{x\to1}7\pi x\cos7\pi x}=\frac1{7\pi\cos7\pi}=-\frac1{7\pi}$

7 0

3 years ago

A theater has 418 seats put in rows of 19 seats each. how many rows are in the theater

GarryVolchara [31]

Answer:

22

Step-by-step explanation:

418 divided by 19 is 22.

8 0

2 years ago

Read 2 more answers

A bank offers a money market account paying 8.7% interest compounded annually. A competing bank offers a money market account p

stich3 [128]

The answer is 8.5% interest compounded daily.

EXPLANATION

Regardless of your rate, the more often interest is paid, the more beneficial the effects of compound interest.

A daily interest account, which has 360 compounding periods a year, in this case, will generate more money than an account with an annual compounding, which has one compounding period per year.

5 0

3 years ago

The increasing annual cost (including tuition, room, board, books, and fees) to attend college has been widely discussed (Time).

NeX [460]

Answer:

(a) PRIVATE COLLEGES

Sample mean is $42.5 thousand

Sample standard deviation is $6.65 thousand

PUBLIC COLLEGES

Sample mean is $22.3 thousand

Sample standard deviation is $4.34 thousand

(b) Point estimate is $20.2 thousand. The mean annual cost to attend private colleges ($42.5 thousand) is more than the mean annual cost to attend public colleges ($22.3 thousand)

(c) 95% confidence interval of the difference between the mean annual cost of attending private and public colleges is $19.2 thousand to $21.2 thousand

Step-by-step explanation:

(a) PRIVATE COLLEGES

Sample mean = Total cost ÷ number of colleges = (51.8+42.2+45+34.3+44+29.6+46.8+36.8+51.5+43) ÷ 10 = 425 ÷ 10 = $42.5 thousand

Sample standard deviation = sqrt[summation (cost - sample mean)^2 ÷ number of colleges] = sqrt([(51.8-42.5)^2 + (42.2-42.5)^2 + (45-42.5)^2 + (34.3-42.5)^2 + (44-42.5)^2 + (29.6-42.5)^2 + (36.8-42.5)^2 + (51.5-42.5)^2 + (43-42.5)^2] ÷ 10) = sqrt (44.24) = $6.65 thousand

PUBLIC COLLEGES

Sample mean = (20.3+22+28.2+15.6+24.1+28.5+22.8+25.8+18.5+25.6+14.4+21.8) ÷ 12 = 267.6 ÷ 12 = $22.3 thousand

Sample standard deviation = sqrt([(20.3-22.3)^2 + (22-22.3)^2 + (28.2-22.3)^2 + (15.6-22.3)^2 + (24.1-22.3)^2 + (28.5-22.3)^2 + (22.8-22.3)^2 + (25.8-22.3)^2 + (18.5-22.3)^2 + (25.6-22.3)^2 + (14.4-22.3)^2 + (21.8-22.3)^2] ÷ 12) = sqrt (18.83) = $4.34 thousand

(b) Point estimate = mean annual cost of attending private colleges - mean annual cost of attending public colleges = $42.5 thousand - $22.3 thousand = $20.2 thousand.

This implies the the mean annual cost of attending private colleges is greater than the mean annual cost of attending public colleges

(c) Confidence Interval = Mean + or - Margin of error (E)

E = t×sd/√n

Mean = $42.5 - $22.3 = $20.2 thousand

sd = $6.65 - $4.34 = $2.31 thousand

n = 10+12 = 22

degree of freedom = 22-2 = 20

t-value corresponding to 20 degrees of freedom and 95% confidence level is 2.086

E = 2.086×$2.31/√22 = $1.0 thousand

Lower bound = Mean - E = $20.2 thousand - $1.0 thousand = $19.2 thousand

Upper bound = Mean + E = $20.2 thousand + $1.0 thousand = $21.2 thousand

95% confidence interval is $19.2 thousand to $21.2 thousand

6 0

3 years ago

Help me now if i get this one wrong i wont be free

jok3333 [9.3K]

Answer: its the first one "opposite sides with equal length"

5 0

3 years ago