1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
GalinKa [24]
3 years ago
12

If it takes 5 elves 5 minutes to make 5 dolls, how long would it take 100 elves to make 100 dolls?

Mathematics
2 answers:
klasskru [66]3 years ago
8 0

Answer: 5 minutes

Step-by-step explanation: If 5 elves can make 5 dolls in 5 minutes then an elf can make 5 dolls in twice the time it would take 5 elves to do it which is 25 minutes.

This means it takes an elf 25/5 = 5 minutes to make one doll

One elf can make 100 dolls in 100 x 5 = 500 minutes

Then it will take 100 elves 500/100 = 5 minutes to make 100 dolls.

NOTE: This method of solving is known as the unitary method. Be careful to note that it won't take 100 elves a hundred minutes to make 100 dolls because it's a concerted effort (each of the 100 elves is making 1 doll in the same time duration (5 minutes))

Natalka [10]3 years ago
7 0
5 elves*5 minutes=5 dolls
1 elf *5 minutes=1 doll

so it takes 5 minutes for an elf to make doll

100 elves*100 minutes/5=100*20=2000 dolls in 100 minutes by 100 elves
You might be interested in
If $525 is 40% of my take home pay for 1 week how much to I make in a week
Vlada [557]

Answer: 525=2/5 so 1/5=262.5 and multiply that by 3 is 787.5 2/5 plus 3/5 is 100% so you would make 787.5

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Question Help Of 515515 samples of seafood purchased from various kinds of food stores in different regions of a country and gen
Setler79 [48]

Answer:

(0.5848 ; 0.6552)

We are confident that about 58% to 66% of sea foods in the country are Mislabelled.

No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.

Step-by-step explanation:

Sample size, n = 51

p = 0.62

1 - p = 1 - 0.62 = 0.38

n = 515

Confidence level = 90% = Zcritical at 90% = 1.645

Confidence interval = (p ± margin of error)

Margin of Error = Zcritical * sqrt[(p(1-p))/n]

Margin of Error = 1.645 * sqrt[(0.62(0.38))/515]

Margin of Error = 1.645 * 0.0214

Margin of Error = 0.035203

Lower boundary = (0.62 - 0.035203) = 0.584797

Upper boundary = (0.62 + 0.035203) = 0.655203

(0.5848 ; 0.6552)

We are confident that about 58% to 66% of sea foods in the country are Mislabelled.

No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.

3 0
3 years ago
Solve for x.
d1i1m1o1n [39]
X³ = 8/125

x = ∛8/125

x = 2/5

In short, Your Answer would be: 2/5

Hope this helps!
5 0
3 years ago
Read 2 more answers
In a recent​ year, a poll asked 2362 random adult citizens of a large country how they rated economic conditions. In the​ poll,
Harman [31]

Answer:

a) The 99% confidence interval is given by (0.198;0.242).

b) Based on the p value obtained and using the significance level assumed \alpha=0.01 we have p_v>\alpha so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

c) \alpha=0.01

Step-by-step explanation:

<em>Data given and notation   </em>

n=2362 represent the random sample taken

X represent the people who says that  they would watch one of the television shows.

\hat p=\frac{X}{n}=0.22 estimated proportion of people rated as​ Excellent/Good economic conditions.

p_o=0.24 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  <em> </em>

<em>Concepts and formulas to use   </em>

We need to conduct a hypothesis in order to test the claim that 24% of people are rated with good economic conditions:  

Null hypothesis:p=0.24  

Alternative hypothesis:p \neq 0.24  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Part a: Test the hypothesis

<em>Check for the assumptions that he sample must satisfy in order to apply the test   </em>

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.  

b) The sample needs to be large enough

np = 2362x0.22=519.64>10 and n(1-p)=2364*(1-0.22)=1843.92>10

Condition satisfied.

<em>Calculate the statistic</em>  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.22 -0.24}{\sqrt{\frac{0.24(1-0.24)}{2362}}}=-2.28

The confidence interval would be given by:

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

The critical value using \alpha=0.01 and \alpha/2 =0.005 would be z_{\alpha/2}=2.58. Replacing the values given we have:

0.22 - (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.198

 0.22 + (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.242

So the 99% confidence interval is given by (0.198;0.242).

Part b

<em>Statistical decision   </em>

P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided is \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z  

So based on the p value obtained and using the significance level assumed \alpha=0.01 we have p_v>\alpha so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

Part c

The confidence level assumed was 99%, so then the signficance is given by \alpha=1-confidence=1-0.99=0.01

6 0
3 years ago
HEY RYAN QUESTIONS COMING SOON!!!!!!!!
Verdich [7]

Answer:

AHHHHH U TELL ME AT THE WORST TIMES HOMIE IMMA TRY THO

8 0
3 years ago
Other questions:
  • Please answer this question!!
    14·1 answer
  • What operation should you use to solve -6
    14·1 answer
  • Equation for (0,6) and (4,5)
    11·1 answer
  • Intersecting lines that form a right angle are defined as _____.
    7·1 answer
  • Which is the graph of linear inequality 6x+2y&gt;-10
    5·1 answer
  • What are the roots of the quadratic function in the graph?
    13·1 answer
  • PLS HELP ME I NEED HELP
    10·1 answer
  • Use the rule 2 + 3 to find the missing number on the table.
    11·1 answer
  • Which numerator makes the statement true enter the correct answer in the box 1/4&lt;5/12&lt; /6?
    15·2 answers
  • 14 less than a number is 2 more than 9 times a number
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!