What is the following quotient? 9+√2/4-√7 Please Help!! thank you.

A company currently has 200 units of a product on hand that it orders every 2 weeks when the salesperson visits the premises. De

disa [49]

Answer: 289 units

Step-by-step explanation:

Given the following :

Inventory (I) = 180

Lead time (L) = 7 days

Review time (T) = 2 weeks = 14 days

Demand (D) = 20

Standard deviation (σ) = 5

Zscore for 95% probability = 1.645

Units to be ordered :

D(T + L) + z(σT+L)

(σT+L) = √(T + L)σ²

= √(14 + 7)5²

= √(21)25

= 22.9

D(T + L) + z(σT+L) - I

20(14 + 7) + 1.645(22.9 + 7) - I

= 420 + 49.1855 - 180

= 289.1855

= 289 quantities

5 0

2 years ago

Little Mexican restaurant sells only two kinds of beef burritos: Mucho beef and Mucho Mucho beef. Last week in the restaurant so

Romashka [77]

Step-by-step explanation:

16+22=38

So then youll check out the price thats worth each. 231$ So what is 38 divided by 231$? Thats your answer

7 0

2 years ago

Jamal works in a city and sometimes takes a taxi to work. The taxicabs charge $1.50 for the first 1/5 mile and $0.25 for each ad

Natalija [7]

1.50+.25x=3.75
.25x=2.25
x=9
9/5+1/5=10/5
2 miles

4 0

2 years ago

Read 2 more answers

Find the sum of the first 63 terms of –19, -13, -7 …

boyakko [2]

The Given Sequence is an Arithmetic Sequence with First term = -19

⇒ a = -19

Second term is -13

We know that Common difference is Difference of second term and first term.

⇒ Common Difference (d) = -13 + 19 = 6

We know that Sum of n terms is given by : $S_n = \frac{n}{2}(2a + (n - 1)d)$

Given n = 63 and we found a = -19 and d = 6

$\implies S_6_3 = \frac{63}{2}(2(-19) + (63 - 1)6)$

$\implies S_6_3 = \frac{63}{2}(-38 + (62)6)$

$\implies S_6_3 = \frac{63}{2}(-38 + 372)$

$\implies S_6_3 = \frac{63}{2}(334)$

$\implies S_6_3 = {63}(167) = 10521$

The Sum of First 63 terms is 10521

4 0

1 year ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$