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olganol [36]
4 years ago
15

HELPPPPP!!!!! Is the following number rational or irrational? \sqrt{99} ​ Choose 1 answer: Choose 1 answer: (Choice A) A Rationa

l (Choice B) B Irrational
Mathematics
1 answer:
Evgen [1.6K]4 years ago
5 0

Answer:

√99 is irrational

Step-by-step explanation:

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if 10 is less than a number is multiplied by 3, the result is equal to 2 times the number. What is the number?
olya-2409 [2.1K]
Hello!

We know that

(x - 10) * 3 = 2x

Distribute the 3

3x - 30 = 2x

Subtract 2x from both sides

x - 30 = 0

Add 30 to both sides

x = 30

The answer is 30

Hope this helps!
4 0
3 years ago
Read 2 more answers
Can you help me plz i will give you brainly its
hodyreva [135]
The answer is 3.5 + 1.5x = 12.5. So the answer would be C.
4 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
3 years ago
How to solve<br> −3−6c=9c+5−15c
Alekssandra [29.7K]
Can salvage 15c+ -15c is 0 + 8 is still 0
7 0
3 years ago
Find the quotient: 7/8  ÷  3/8<br><br> A 21/64<br> B 3/7<br> C 2 1/3<br> D 7 1/3
pantera1 [17]
D

explanation:

7/8 x 8/3=56/24—simplify—-> 7/3
4 0
3 years ago
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