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marysya [2.9K]
2 years ago
13

10. +

Mathematics
1 answer:
Sidana [21]2 years ago
3 0

Answer:

12 boxes or C.

Step-by-step explanation:

Max = divide.

GCF=12.

LCM = 144.

Difference:

144/12 or 12.

So 12 boxes

Hope this helps :D

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I have seen two answers for this and i want a third opinion, PlEaSe help me!!A man selling hotdogs, french fries, and a soda as
irina1246 [14]

Answer:

50 combos

Step-by-step explanation:

Put it into an equation first in order to find x.

2x + 250 = 7x

Subtract 2x from both sides.

250 = 5x

Then divide both sides by 5.

x = 50

Therefore you need 50 combo meals in order to make the amount made equal to the amount paid.

<u>You can check your work by plugging in the number to both.</u>

$250 plus the $2 times 50 combo meals.

250 + 2(50) = 350

$7 times 50 combo meals

7 x 300 = 350

8 0
3 years ago
HELP MEH NOW PLZ :_(
bulgar [2K]

Answer: hey sup fellow player

Step-by-step explanation: how’s life eh?

5 0
3 years ago
Last year, Justin opened an investment account with $5400. At the end of the year, the amount in the account had increased by 6.
lara31 [8.8K]

Answer:

$351 and $5751

Step-by-step explanation:

Given, Justin opened an investment account with $5400. At the end of the year, 6.5% increased. So, the total increase in the dollar is 5400* 0.065 = 351. The increase in the dollar is $364.

At the end of the year, Justin will have 5400 + 351 = $5,751. So, Justin will have $5,751 at the end of the year in his investment bank account.

6 0
3 years ago
Read 2 more answers
Ann has 30 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variable
IgorLugansk [536]

Answer:

P(T_A < T_B) = P(T_A -T_B

Step-by-step explanation:

Assuming this problem: "Ann has 30 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 50 minutes and standard deviation 10 minutes. Bob has 30 jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with mean 52 minutes and standard deviation 15 minutes. Ann's and Bob's times are independent. Find the approximate probability that Ann finishes her jobs before Bob finishes his jobs".

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

For this case we can create some notation.

Let A the values for Ann we know that n1 = 30 jobs solved in sequence and we can assume that the random variable X_i the time in order to do the ith job for i=1,2,....,n_1. We will have the following parameters for A.

\mu_A = 50, \sigma_A =10

W can assume that B represent Bob we know that n2 = 30 jobs solved in sequence and we can assume that the random variable [tex[X_i[/tex] the time in order to do the ith job for i=1,2,....,n_2. We will have the following parameters for A

\mu_B = 52, \sigma_B =15

And we can find the distribution for the total, if we remember the definition of mean we have:

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

And T =n \bar X

And the E(T) = n \mu

Var(T) = n^2 \frac{\sigma^2}{n}=n\sigma^2

So then we have:

E(T_A)=30*50 =1500 , Var(T_A) = 30*10^2 =3000

E(T_B)=30*52 =1560 , Var(T_B) = 30 *15^2 =6750

Since we want this probability "Find the approximate probability that Ann finishes her jobs before Bob finishes his jobs" we can express like this:

P(T_A < T_B) = P(T_A -T_B

Since we have independence (condition given by the problem) we can find the parameters for the random variable T_A -T_B

E[T_A -T_B] = E(T_A) -E(T_B)=1500-1560=-60

Var[T_A -T_B]= Var(T_A)+Var(T_B) =3000+6750=9750

And now we can find the probability like this:

P(T_A < T_B) = P(T_A -T_B

P(\frac{(T_A -T_B)-(-60)}{\sqrt{9750}}< \frac{60}{\sqrt{9750}})

P(Z

7 0
3 years ago
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faust18 [17]
I pretty sure b
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