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Sonbull [250]
3 years ago
5

WILL GIVE BRAINLIEST!!!!!!! A boy had 20 cents. He bought x pencils for 3 cents each. If y equals the number of pennies left, wr

ite an equation showing the dependence of y on x. What is the domain of the function?But it says that x is a whole number for the domain. PLS HELP.
Mathematics
1 answer:
kumpel [21]3 years ago
5 0

Answer:

Hey there!!

The total number of cents - 20

Cost for each pencil - 3 cents

Number of pencils bought - ' x '

y = number of pennies left

What is the domain ?

Show how y is dependent on x ..

Let's get this into an equation :

... The total cost for x pencils bought = 3x

... Number of pennies left = 20 - 3x

... y = number of pennies left

... y = 20 - 3x

Notice : If the x value changes, the y value changes too

... If x = 1 , then , y = 17; If x = 2 , then , y = 14

Hence, we could say y is dependent upon x

Domain = ?

... Remember - The total number of pennies = 20 ; hence, the total cost cannot go above 20 cents.

Hence, we will have to work with inequalities

The equation :

... 20 ≥ 3x

Divide 3 on both sides

20 / 3 ≥ x

x ≥ 6.667

Let's take this as 6

Hence , the domain will be :

D : { 1 , 2 , 3 , 4 , 5 , 6 }

Hope my answer helps!

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Which of the following is determined as a direct result of computing the earliest starting and finishing times for the activitie
Lady bird [3.3K]

Answer:

Expected project duration  is determined as a direct result of computing the earliest starting and finishing times for the activities of a project network.

Step-by-step explanation:

In project management, early start is the earliest moment at which a specific activity may start and early finish the earliest moment at which it can end. Early finish (es)  is computed as:

EF = ES + d

where d is the activity duration and ES early start

Early start is computed as

ES: Max (EF-1)

or the maximum early start of predecessory activities.

Once we have calculated these values in the network for each activity , the early finish of last activity corresponds to the expected project duration, the earliest time in which we may finish the project if there are no issues.

Max project duration is open question as we could have infinite delays or never finish the project. Time variance in project duration may only be estimated once we have actual execution times of our project.

Slack time and critical path are obtained after obtaining the ES and EF but this info alone is not sufficient. We require either the late start or the late finish of the activities to calculate slack, Zaero slack activities, those that cannot be delayed form critical path and can only be obtained after having ES. EF. LS and LF

5 0
3 years ago
PLEASE HELP! IF CORRECT WILL GET BRAINLIST!
viva [34]

Answer:

2

Step-by-step explanation:

f=1

2 x 1=2

4-2=2

7 0
4 years ago
Read 2 more answers
Someone help me pls. I’ll mark u BRAINLIEST
Kobotan [32]

Answer:

aight what's the question

7 0
3 years ago
Rauls car averages 17.3 miles per gallon of gasoline. How many miles can Raul drive if he fills his tank with 10.5 gallons of ga
Lisa [10]

Answer:

181.65 miles for 10.5 gallons of gas

Step-by-step explanation:

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4 years ago
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According to a survey done at your school, about 42% of all the female students participate in 2 sport seasons. You randomly ask
IRISSAK [1]

Answer: 0.353

Step-by-step explanation:

We know that 42% of all the female students participate in 2 sports this season. (58% is the percent that did not)

This means that if we select a random student, there is a 0.42 probability that the student did participate in 2 sports this season.

Other thing that i will use is that, if we have a group of N objects, and we want to create a group of K objects out of the N, the number of combinations is:

c = \frac{N!}{(N - K)!*K!}

Now, out of 5, we want to find the proability that, at least 3 of them, did participate in at least 2 sports this season.

The probability that all 5 of them participated, is equal to the product of the individual probabilities. (so N = 5, and K = 5)

P = (0.42^5)*1 = 0.013

Now, if only 4 of them did participate (N = 5, k = 4), the probability is equal to:

P = (0.42^4)*(0.58)*5!/4! = (0.42^4)*(0.58)*5 = 0.09

The probability for only 3 is: (N = 5 and K = 3)

P = (0.42^3)*(0.58^2)*5!/(3!*2!) = (0.42^3)*(0.58^2)*(5*4/2) = 0.25

Then the total probability is:

P = 0.013 + 0.09 + 0.25 = 0.353

7 0
3 years ago
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