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nikdorinn [45]
2 years ago
15

4−(−8)= please help i need it now

Mathematics
2 answers:
Jlenok [28]2 years ago
6 0

Answer:

Step-by-step explanation:

s344n2d4d5 [400]2 years ago
4 0

Answer:

hope it helps...

- multiplied with another - gives us a +.

While opening a bracket we consider a sign of multiplication. so the equation becomes 4 + 8 which gives us 12. Thus 12 is the answer.

pls give brainliest for the answer

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Imagine you have some workers and some handheld computers that you can use to take inventory at a warehouse. There are diminishi
Svet_ta [14]

Answer:

a. $1.03

b. $0.93

c. 0.98

d. 2 workers

Step-by-step explanation:

a. Given that:

  • 1 computer : 1 worker : inventory 150 items per hour
  • 1 computer : 2 workers : inventory 200 items per hour
  • 1 computer : 3 workers : inventory 220 items per hour
  • 1 computer : 4+ workers : fewer than 235 items per hour
  • Cost: $100 per computer ; $25 per worker

The fixed production factor in the warehouse is the computer used:

-One computer used, but the number of users is varied to inventory a specified number of items.

-The variable production factor is the number of workers assigned per one computer.

#The cost of inventorying a single item by one worker is:

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_1=\frac{125+30}{150}\\\\\\=1.03

Hence, the cost of inventorying a single item is $1.03

b. Using the information provided above, the cost of inventorying a single item when two workers are assigned is :

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_2=\frac{125+2\times30}{200}\\\\\\=0.925

Hence, the cost of inventorying a single item is $0.93

c.Using the information provided above, the cost of inventorying a single item when three workers are assigned is :

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_3=\frac{125+3\times30}{220}\\\\\\=0.98

Hence, the cost of inventorying a single item is $0.98

d. To determine the most cost-effective job assignment, we calculate the cost of 4+ workers.

Take any number less than 235(say 234) as the inventory units:

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_4=\frac{125+4\times30}{234}\\\\\\=1.05

From our calculations, it's clear that two workers per computer costs the least amount($0.93) per unit item. Hence, it is best to assign two workers per computer.

4 0
3 years ago
Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8
Delvig [45]
We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34
Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86

Thus the 98% confidence interval will be (105.34, 141.86)


Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.
3 0
3 years ago
How many cubic feet of soil are needed to fill a raised vegetable bed that measure 12 feet by 10 feet by 9 inches?
otez555 [7]

18 feet 1 feet is equivalent to 12 inches

3 0
3 years ago
arts academy requires there to be 6 teachers for every 144 students and 3 tutors for every 24 students. How many students does t
jonny [76]

Answer:

the teachers would have 24 students per teacher. 10 students per tutor. There would need to be 11 tutors

Step-by-step explanation:

each tutor would have 10 students then if you do 11 times 10 you get 11

(hope this helps can I pls have brainlist (crown)☺️)

4 0
2 years ago
A survey of students at a particular college showed that 33% of them prefer the American Heritage Dictionary (as opposed to Webs
Juliette [100K]

Answer:

5.13 % to the nearest hundredth.

Step-by-step explanation:

This is a Binomial Distribution .

The required probability

= 30C6 (0.33)^6 (0.67)^(30-6)

= 0.051345

7 0
2 years ago
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