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deff fn [24]
3 years ago
12

LOTS OF POINTS!! A cyclist rides a 25 mile stretch of road at a constant speed. It takes her 90 minutes to ride the 25 miles . A

t what speed is she riding? Please tell me the correct answer and how you got the answer step by step!!! Whoever does this will get the brainliest!!!
Mathematics
1 answer:
sattari [20]3 years ago
7 0
25m/90minutes = .278 mile/minute

Divide 25/90 to get .278

Hope that helps
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If x=4cos(2t) and y=4sin(2t), what is the parameter of these equations
kobusy [5.1K]

Answer:

y=√(4+x)(4−x)

Step-by-step explanation:

7 0
2 years ago
Box A contains 2 red and 2 green balls and box B contains 1 red and 2 green balls. A box is selected at random and then a ball i
serious [3.7K]
The answer is 1/2. I think that since the question is about Box B , only boxes will be taken into consideration. Since the total number of boxes are two and box b is one ,the probability of the box being B is 1/2. 
5 0
3 years ago
A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or
Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the

denote the event that he passes the examination. Then,

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

The events (Y|X) follows a Binomial distribution with probability of success 0.80 and the events (Y|X^{c}) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

P(X)=2\cdot P(X^{c})

Then,

P(X)+P(X^{c})=1

⇒

2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}

Then,

P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}

Compute the probability that the students passes if request an examination with 3 examiners as follows:

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

        =[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}

       =0.715

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

        =[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}

       =0.734

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0
2 years ago
Can i get help on question 7 please?
CaHeK987 [17]

Do you have notes so I can look at them to help you? Cause I pretty good at math but I need the notes.


3 0
3 years ago
Omar had 7 goldfish in his fish tank. He has 3 fewer redfish than goldfish in his tank. Omar also has 8 more black fish than gol
Katarina [22]

Answer: 26

Step-by-step explanation: he will have 4 red fish 15 black fish then you add them together 4+7=11 11+15=26

6 0
3 years ago
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