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o-na [289]
3 years ago
14

Increase $200 by 50%. PLEASE HELP ME!

Mathematics
1 answer:
IRINA_888 [86]3 years ago
5 0

It would be: $200 + (200*50)/100

             =200 + 100

             =$300.00

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Why is the number of operators per shift multiplied by approximately 4.5 to obtain the total number of operators required to run
4vir4ik [10]

I believe you meant "why is the number of shifts multiplied by approximately 4.5 to obtain the total number of operators required to run the plant"

Answer and Explanation:

There are 3 shifts per day, 49 weeks per year and 5 shifts per operator per week

To get total number of operators required to run the plant, we multiply number of shifts in a year by number if operators per shift.

49 weeks×5 shifts= 245 shifts per operator per year

365×3 shifts= 1095 shifts per year

1095/245=4.5 operators per shift

total number of operators required to run the plant(per day) = 4.5×3= 13.5 approximately 14

total number of operators required to run the plant(per year) =4.5×1095=4927.5 approximately 4928

3 0
3 years ago
HELP 25pts and will mark brainliest. I NEED THIS NOW!! Which expression shows the result of applying the distributive property t
Arte-miy333 [17]
The last bottom problem
6 0
3 years ago
The accompanying data came from a study of collusion in bidding within the construction industry. No. Bidders No. Contracts 2 6
viktelen [127]

Answer:

a)

The proportion of contracts involved at most five bidders is 0.667.

The proportion of contracts involved at least five bidders is 0.51.

b)

The  proportion of contracts involved  between five and 10 inclusive bidders is 0.5.

The  proportion of contracts involved  strictly between five and 10 bidders is 0.304.

Step-by-step explanation:

No. bidders    No. contracts     Relative frequency of contracts

2                         6                       6/102=0.0588

3                         20                     20/102=0.1961

4                         24                     24/102=0.2353

5                         18                       18/102=0.1765

6                         13                        13/102=0.1275

7                          7                          7/102=0.0686

8                          5                          5/102=0.049

9                          6                          6/102=0.0588

10                         2                           2/102=0.0196

11                          1                            1/102=0.0098

Total                  102

a)

We have to find proportion of contracts involved at most five bidders.

Proportion of at most 5= Relative frequency 2+ Relative frequency 3+ Relative frequency 4+ Relative frequency 5

Proportion of at most 5=0.0588+ 0.1961+0.2353+0.1765

Proportion of at most 5=0.6667

The proportion of contracts involved at most five bidders is 0.667.

proportion of at least five bidders= proportion≥5= 1- proportion less than 5

Proportion less than 5=0.0588+ 0.1961+0.2353=0.4902

proportion of at least five bidders=1-0.4902=0.5098

The proportion of contracts involved at least five bidders is 0.51

b)

We have to find proportion of contracts involved  between five and 10 inclusive bidders.

Proportion of contracts between five and 10 inclusive= Relative frequency 5+ Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9+ Relative frequency 10

Proportion of between five and 10 inclusive=0.1765 +0.1275 +0.0686 +0.049 +0.0588 +0.0196

Proportion of between five and 10 inclusive=0.5

The  proportion of contracts involved  between five and 10 inclusive bidders is 0.5

We have to find proportion of contracts involved  strictly between five and 10 bidders.

Proportion of contracts strictly between five and 10=  Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9

Proportion of strictly between five and 10=0.1275 +0.0686 +0.049 +0.0588

Proportion of strictly between five and 10=0.3039

The  proportion of contracts involved  strictly between five and 10 bidders is 0.304.

8 0
3 years ago
Find the value of x to the nearest tenth tan x =5
tankabanditka [31]

Answer:

x ≈ 78.7° (in degree)

x ≈ 1.4      (in radians)

Step-by-step explanation:

Given in the question an equation

tan(x) = 5

       x =  tan^{1}(5)

       x = 78.69

       x ≈ 78.7°

In radian:

x ≈ 1.4

4 0
3 years ago
Read 2 more answers
Anybody knows the answer?
soldi70 [24.7K]

Hey!


RS = 15 and RT = 18. Subtract 18 from 15 to get your answer.


18 - 15 = 3


<em>That means your answer is 3</em>


\framebox{Answer = 3}

8 0
2 years ago
Read 2 more answers
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