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Mars2501 [29]
3 years ago
12

It costs $3.45 to buy 1/2 lb. of chopped walnuts. How much would it cost to purchase 7 lbs. of walnuts?

Mathematics
2 answers:
Rasek [7]3 years ago
8 0

Answer:

3.45/0.5= 6.9

6.9*7=48.3

Step-by-step explanation:

e-lub [12.9K]3 years ago
5 0

Answer:It would cost $48.3

Step-by-step explanation:7 lb is 14 times 1/2 lb. If you want to figure out the price youd multiply 3.45 by 14 as well. Youd get 48.3 in that case

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He arrives at 9:30 because if you take 8:40 (when dealing with time I like to think of it as a whole number so 8:40 would become 840) then just add 50 minutes to 8:40 (or 840+50 which equals 890 also know as 9:30 because you have to subtract 60 from 90 (because 60 is the limit in time (well technically 59 is)) then add one to 8 like so, which equals 9:30 Enjoy!=)
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Below are the steps to solve an equation:
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A recycling truck begins its weekly route at the recycling plant at point A, as pictured on the coordinate plane below. It trave
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3 years ago
What's equivalent to 2.5(4.2+7.4y)-3.6(8x-4.5y)-(2.4x+6y)
irina1246 [14]

Answer:

-31.2x + 28.7y + 10.5

Step-by-step explanation:

Step 1: Write expression

2.5(4.2 + 7.4y) - 3.6(8x - 4.5y) - (2.4x + 6y)

Step 2: Distribute

10.5 + 18.5y - 28.8x + 16.2y - 2.4x - 6y

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5 0
3 years ago
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

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4 years ago
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