After applying the composition / to the point (–5, 3), in which quadrant does the image lie?

A product with an annual demand of 1000 units has Co = $25.50 and Ch = $8. The demand exhibits some variability such that the le

frosja888 [35]

Answer:

a) Recommended order quantity = 79.84 = 80

b) recorder point = 35

Safety stock = 10

Safety stock cost = $0/year

ci) P(stock out/cycle) = 0.1587

cii) Number of stock outs/ year = 2

Step-by-step explanation:

The pictures attached show a clear explanation of all the processes.

7 0

3 years ago

According to a study conducted by the Toronto-based social media analytics firm Sysomos, of all tweets get no reaction. That is,

amm1812

Complete Question

According to a study conducted by the Toronto-based social media analytics firm Sysomos, 75% of all tweets get no reaction. That is, these are tweets that are not replied to or retweeted (Sysomos website, January ). Suppose we randomly select tweets.

a. What is the expected number of these tweets with no reaction (to the nearest whole number)?

b. What are the variance (to 2 decimals) and standard deviation (to 4 decimals) for the number of these tweets with no reaction? Variance Standard deviation

Answer:

a

The mean is $\mu = 150$

The variance $\sigma^2 =37.5$

The standard deviation is $\sigma \approx 6.1237$

Step-by-step explanation:

A binomial distribution is mathematically represented as $B(n,p)$

where n denotes the number of independent trials =200

And p denote the probability of success = 0.75

We also have the probability of failure which is denoted as q and is equivalent to $1-p$

if p and q are constant then

In binomial distribution the probability of x success and (n-x) is mathematically represented as

$P[X =x] =[\left n} \atop x}} \right. ] p^x (1-p)^{n-x}$

Where x= 0,1,2,...,n $for 0\le p \le 1$

Now $[\left n} \atop x}} \right. ] = \frac{n!}{x(n-x)!}$

Now the mean(Expected number ) of these probability in binomial distribution is mathematically represented as

$\mu = np$

$= 200 *0.75 = 150$

And the variance is mathematically represented as

$\sigma^2 = np(1-p)$

$= 200 * 0.75 (1-0.75) = 37.5$

And the standard deviation is mathematically represented as

$\sigma = \sqrt{np(1-p)}$

$= \sqrt{200 *0.75(1-0.75)} =\sqrt{37.5} \approx 6.1237$

We can obtain the standard z -score of a d

6 0

3 years ago

use the slope formula to determine the slope of a line that passes through the points (4,9) and (-8,-6)

larisa86 [58]

$\frac{-6-9}{-8-4}\\ \\\frac{-15}{-12}\\ \\= \frac{5}{4}$

Slope: 5/4

6 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

4/5 of 40

algol [13]

4/5 of 40 = 32
1/3 of 3 = 1
1/3 of 15 muffins = 10 muffins

3 0

3 years ago