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Rama09 [41]
3 years ago
7

Eric runs a race that can be modeled by the equation shown , where d is his distance, in feet , from the starting line and t is

his time, in seconds, since the race began .
d=8t-4
Which equation shows Eric’s time of distance?

Mathematics
2 answers:
gladu [14]3 years ago
4 0

Answer:D+4/8

Step-by-step explanation:

slamgirl [31]3 years ago
3 0

Answer:

C.  t = (d + 4)/8

Step-by-step explanation:

The equations showing Eric time of distance simply means making t the subject of the formula in the equation given below.

d = 8t - 4

Transfer -4 to the left side and it becomes positive

8t = d + 4

divide both sides by 8

t = (d + 4)/8

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Arlecino [84]

Answer:

First answer is 127 and second answer is 15

8 0
2 years ago
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Select the correct answer.
valentinak56 [21]

A

see attached for explanation

I hope it helps

7 0
2 years ago
A tortilla maker produces 100 tortillas in half an hour. if he works from 9 am to 5 pm with a one hour lunch and two 15-minute b
ira [324]
The answer is 1300 i think,
hope it helps!
3 0
3 years ago
Read 2 more answers
A man notices that a tank is half full. After emptying 600 litres from the tank, that it is now one-third full. How much does th
enot [183]
One half is equal to 3/6.
One third is equal to 2/6.

After removing 600 litres from the tank, there is a decrease from 3/6 to 2/6

3/6-2/6=1/6

600 litres is 1/6 of the tank. Multiply by 6 to get the capacity of a full tank.

600*6=3600

The tank can hold 3600 litres.
6 0
3 years ago
The probability that two people have the same birthday in a room of 20 people is about 41.1%. It turns out that
salantis [7]

Answer:

a) Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Solution to the problem

Part a

Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

Part b

We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

4 0
3 years ago
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