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Vlad1618 [11]
2 years ago
12

At the Hula Factory, the assembly line produces 212 leis in a batch. The leis are packaged in boxes with 30 each. If 14 batches

are made in one day, how many leis will be leftover at the end of it?
Mathematics
1 answer:
nikklg [1K]2 years ago
7 0

Answer:

197 leis left over

Step-by-step explanation:

14 times 212 = 2968

2968/30 = 98.93

212 times 0.93 = 197

98 packages and 197 leis left over

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Please give me the correct answer ​
sergejj [24]

Answer:

16 units

Step-by-step explanation:

6.1 + 5.8 + 4.1 = 16

5 0
3 years ago
Since 2010, when 102390 Cases were reported, each year the number of new flu cases decrease to 85% of the prior year. Predict th
BaLLatris [955]

Answer:

20,158 cases

Step-by-step explanation:

Let t=0 represent year 2010.

We have been given that since 2010, when 102390 Cases were reported, each year the number of new flu cases decrease to 85% of the prior year.

Since the flu cases decrease to 85% of the prior year, so the flu cases for every next year will be 85% of last year and decay rate is 15%.

We can represent this information in an exponential decay function as:

F(t)=102,390(1-0.15)^t

F(t)=102,390(0.85)^t

To find number of cases in 2020, we will substitute t=10 in our decay function as:

F(10)=102,390(0.85)^{10}

F(10)=102,390(0.1968744043407227)

F(10)=20,157.970260446597\approx 20,158

Therefore, 20,158 cases  will be reported in 2020.

3 0
3 years ago
Provided below are summary statistics for independent simple random samples from two populations. Use the nonpooled​ t-test and
Vesna [10]

Answer:

okay not clear,

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5 0
3 years ago
Find the sum 1/6 (12C + 24) + 1/3(12- 3)<br><br>a-b-c- or d
Anarel [89]
<span>1/6 (12C + 24) + 1/3(12 c -  3) = 

12 c/ 6  + 4 +12c / 3 - 1 = 

2c +4 + 4 c -1 

6c +3 </span>
7 0
3 years ago
You are working in a small, student-run company that sends out merchandise with university branding to alumni around the world.
topjm [15]

Answer:

The answer is "0.142466".

Step-by-step explanation:

Using the p formula for the proportion of nonconforming units through the subgrouping which can vary in sizes:

p =\frac{np}{n}\\\\

\bar{p}=\frac{\Sigma np}{\Sigma n}\\\\

Defects = \frac{5}{100} \times 50 \\\\

p = \frac{5}{100} \times \frac{50}{50}=0.05\\\\

It calculates the controls limits through the p-chart that is:

UCL_{p},LCL_{p}=\bar{p} \pm \sqrt{\frac{\bar{p}(1-\bar{p})}{\bar{n}}}\\\\

So, the upper control limits:

= 0.05 + 3 \times \sqrt{(\frac{0.05\times(1-0.05)}{50})} \\\\= 0.142466

7 0
3 years ago
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