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Mekhanik [1.2K]
2 years ago
11

A crew will arrive in one week and begin filming a city for a movie. The mayor is desperate to clean the city streets before fil

ming begins. Two teams are​ available, one that requires 700 hours and one that requires 600 hours. If the teams work​ together, how long will it take to clean all of the​ streets? Is this enough time before the cameras begin​ rolling?
Mathematics
1 answer:
Reptile [31]2 years ago
4 0

Step-by-step explanation:

1 week = 7 days of 24 hours each = 7×24 = 168 hours.

the first team can do 1/700 of the total work per hour.

the second team can do 1/600 if the total work per hour.

if they work together, then 1/700 + 1/600 of the total work can be done per hour.

to add these 2 fractions we need to bring them to the same denominator. and that would be 4200 (LCM of 700 and 600).

1/700 + 1/600 = 6/4200 + 7/4200 = 13/4200

so, together, they can do 13/4200 if the total job in 1 hour.

how many hours will they need together to finish the job ?

well, how often does 13 fit into 4200 (as 4200/4200 is the complete job) ?

4200 / 13 = 323.0769231... hours

so, it will still take a little bit over 323 hours to clean all the streets.

but there are only 168 hours until the film crew starts.

so, no, there is not enough time.

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Answer:

3.44 ounces in each bowl

Step-by-step explanation:

15.34 ounces total

15.34-1.58=13.76 (ounces in all 4 bowls)

13.76/4=3.44 ounces in each bowl

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3 years ago
1) Lithium isotope rations are important to medicine, the 6Li/7Li ratio in a standard reference material was measured several ti
uysha [10]

Answer:

1) 0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588    

0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219    

b) ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653

And we want 2/3 of the margin of error so then would be: 2/3 ME = 0.00001111

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (2)

Replacing we got:

n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12

So the answer for this case would be n=12 rounded up to the nearest integer

Step-by-step explanation:

Information given

0.082601, 0.082621, 0.082589, 0.082617, 0.082598

We can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=0.0826052 represent the sample mean

\mu population mean

s=0.000013424 represent the sample standard deviation

n=5 represent the sample size  

Part 1

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The degrees of freedom, given by:

df=n-1=5-1=4

The Confidence level is 0.95 or 95%, and the significance would be \alpha=0.05 and \alpha/2 =0.025, the critical value would be using the t distribution with 4 degrees of freedom: t_{\alpha/2}=2.776

Now we have everything in order to replace into formula (1):

0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588    

0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219    

Part 2

The original margin of error is given by:

ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653

And we want 2/3 of the margin of error so then would be: 2/3 ME = 0.00001111

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (2)

Replacing we got:

n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12

So the answer for this case would be n=12 rounded up to the nearest integer

3 0
3 years ago
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Svetradugi [14.3K]

Answer:

Ryan in 1 hour can code 4/5 of his game

Step-by-step explanation:

we know that

In 3/4 of an hour Ryan codes 3/5 of his game

so

using proportion

Find out how much of the game can he code in 1 hour

Let

x ----> fraction of the game that Ryan can code in 1 hour

\frac{(3/4)}{(3/5)}\frac{hours}{code} =\frac{1}{x}\frac{hour}{code} \\\\x=(3/5)/(3/4)\\\\x=\frac{12}{15}\ code

Simplify

x=\frac{4}{5}\ code

therefore

Ryan in 1 hour can code 4/5 of his game

5 0
3 years ago
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