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gulaghasi [49]
2 years ago
13

Dina and Masha started out on a 20 miles bike path at the same time. When Dina reached the end of the 20 miles, Masha still had

4 miles left to bike. If Dina’s biking speed was 10 mph, find Masha’s biking speed.
Mathematics
2 answers:
leonid [27]2 years ago
4 0

Answer:

8mph

Step-by-step explanation:

(20-4)/2

=16/2

=8

Bingel [31]2 years ago
3 0

Answer:

8

Step-by-step explanation:

(20-4)/2=16/2=8

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(Pts: 1) Please help me! Question is in attachment!
Sloan [31]
- - Solve:

To find the volume of a sphere:
 \frac{4}{3} *3.14*R^3

3.14≈π

Now since its telling us to use π, then we do our formula by multiplying by π.

π=3.14 (Approximate amount)

So:

4/3*3.14*9^3=3052.08

So you are correct.

When Radius=9, you plug in for R. :)

8 0
3 years ago
Read 2 more answers
Solve for x. <br><br> 5x - 1 = 26<br><br> x = 27/5<br> x = 5<br> x = -5
yan [13]
5x = 26 + 1
5x = 27
x = 27/5

Answer: A) or the first option ✅
4 0
3 years ago
Brad has two alloys containing copper and silver. Alloy A contains 50% pure silver, and alloy B contains 75% pure silver. He wan
oee [108]

Answer:

I think the answer would be

Alloy A 30%

Alloy B 40%

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8 0
3 years ago
Read 2 more answers
- Which is greater?<br> 5 2<br> a) or<br> 16 9
azamat

Answer:

5/2

Step-by-step explanation:

5/2= 2 1/2

while 16/9=1 7/9

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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