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2 years ago

14

This time the florist is making bouquets with daisies and carnations. She has 35 daisies and 42 carnations. She wants to put an

equal amount of each kind of flower in the bouquets. What is the greatest number of bouquets the florist can make

1 answer:

mash [69]2 years ago

6 0

Answer:

7 flowers

Step-by-step explanation:divide 7 from each and you'll gat your answer

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Simplify (4x + 12) - (6x - 18)

bixtya [17]

Answer:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

4*x-12-(6*x-18)=0

Step-by-step explanation:

Pull out like factors :

-2x + 6 = -2 • (x - 3)

5 0

2 years ago

Read 2 more answers

In the book Business Research Methods, Donald R. Cooper and C. William Emory (1995) discuss a manager who wishes to compare the

S_A_V [24]

Answer:

Null hypothesis is: U1 - U2 ≤ 0

Alternative hypothesis is U1 - U2 > 0

Step-by-step explanation:

The question involves a comparison of the two types of training given to the salespeople. The requirement is to set up the hypothesis that type A training leads to higher mean weakly sales compared to type B training.

Let U1 = mean sales by type A trainees

Let U2 = mean sales by type B trainees

Therefore, the null hypothesis (H0) is: U1 - U2 ≤ 0

This implies that type A training does not result in higher mean weekly sales than type B training.

The alternative hypothesis (H1) is: U1 - U2 > 0

This implies that type A training indeed results in higher mean weekly sales than type B training.

4 0

3 years ago

For the standard normal curve, find the z-score that corresponds to the 30th percentile. -0.47 -0.98 -0.53 -0.12

Nata [24]

We are looking for the Zo score that corresponds to the 30th percentile.
We use the table for the standard normal distribution.
From P (Z <Zo) = 0.3 we search for Zo.
Then we look in the table for the value of Zo that corresponds to 0.3.
Hence, the approximate value is -0.53.
The option is the third

7 0

3 years ago

Assume a warehouse operates 24 hours a day. Truck arrivals follow Poisson distribution with a mean rate of 36 per day and servic

kirill [66]

The expected waiting time in system for typical truck is 2 hours.

Step-by-step explanation:

Data Given are as follows.

Truck arrival rate is given by, α = 36 / day

Truck operation departure rate is given, β= 48 / day

A constructed queuing model is such that so that queue lengths and waiting time can be predicted.

In queuing theory, we have to achieve economic balance between number of customers arriving into system and that of leaving the system whether referring to people or things, in correlating such variables as how customers arrive, how service meets their requirements, average service time and extent of variations, and idle time.

This problem is solved by using concept of Single Channel Arrival with exponential service infinite populate model.

Waiting time in system is given by,

$w_{s} = \frac{1}{\alpha - \beta }$

where $w_s$ is waiting time in system

$\alpha$ is arrival rate described Poission distribution

$\beta$ is service rate described by Exponential distribution

$w_{s} = \frac{1}{\alpha - \beta }$

$w_{s} = \frac{1}{48 - 36 }$

$w_{s} = \frac{1}{12 } day$

$w_{s} = \frac{1}{12 } \times 24 hour$ ...it is due to 1 day = 24 hours

$w_{s} = 2 hours$

Therefore, time required for waiting in system is 2 hours.

5 0

3 years ago

4) BRAINLIEST + 15 + POINTS :)

malfutka [58]

Answer:

C

Step-by-step explanation:

New equation: y = -3(2y + 3) - 4

Distribute and simplify: y = -6y - 13

Add 6y: 7y = -13

6 0

3 years ago