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3 years ago

If there are five choices on a multiple choice test, what is the probability of guessing any of them correctly?

1 answer:

4 0

C. 1/5. Because there is only one answer that is the right answer to each multiple choice. Also, there are 5 choices for us to choose.

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Factorize this expression completely <br>using (a-b)² formula<br>3a²- 36a + 108

Novosadov [1.4K]

Answer:

$3a^2 - 36a + 108 = (\sqrt{3}a - 6\sqrt{3} )^2$

Step-by-step explanation:

$3a^2 = (\sqrt{3} a)^2\\108 = 9 \times 4 \times 3= 3^2 \times 2^2 \times 3 = (3\times2 \sqrt{3} )^2 = (6\sqrt{3})^2$

$so, a = \sqrt{3} a \ , \ b = 6\sqrt{3} \\\\(a - b)^2 = a^2 -2ab + b^2\\\\-36a = -2ab\\\\-36a = -2 \times \sqrt{3} a \times 6\sqrt{3}$

Therefore ,

$3a^2 - 36a + 108 = (\sqrt{3}a )^2 - 2(\sqrt{3}a \times 6\sqrt{3} ) + (6\sqrt{3} )^2 = (\sqrt{3}a - 6\sqrt{3} )^2$

6 0

3 years ago

Find the length between (1, 3) and (5, 6)

ipn [44]

Answer:

Step-by-step explanation:

6 0

3 years ago

Five males with an X-linked genetic disorder have one child each. The random variable x is the number of children among the fiv

Dahasolnce [82]

Answer and Explanation:

Given : Five males with an X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the X-linked genetic disorder.

To find :

a. Does the table show a probability distribution?

b. Find the mean and standard deviation of the random variable x.

Solution :

a) To determine that table shows a probability distribution we add up all six probabilities if the sum is 1 then it is a valid distribution.

$\sum P(X)=0.029+0.147+0.324+0.324+0.147+0.029$

$\sum P(X)=1$

Yes it is a probability distribution.

b) First we create the table as per requirements,

x P(x) xP(x) x² x²P(x)

0 0.029 0 0 0

1 0.147 0.147 1 0.147

2 0.324 0.648 4 1.296

3 0.324 0.972 9 2.916

4 0.147 0.588 16 2.352

5 0.029 0.145 25 0.725

∑P(x)=1 ∑xP(x)=2.5 ∑x²P(x)=7.436

The mean of the random variable is

$\mu=\sum xP(x)=2.5$

The standard deviation of the random sample is

$Variance=\sigma^2$

$\sigma^2=\sum x^2P(x)-\mu^2$

$\sigma^2=7.436-(2.5)^2$

$\sigma^2=7.436-6.25$

$\sigma^2=1.186$

$\sigma=\sqrt{1.186}$

$\sigma=1.08$

Therefore, The mean is 2.5 and the standard deviation is 1.08.

5 0

3 years ago

Find the domain for the expression:<br> y/y-5 + 1/y+8

Jet001 [13]

Answer:

1/y+4

Step-by-step explanation:

1-5+1/y+8

1/y+(1-5+8)

1/y+4

7 0

3 years ago

Read 2 more answers

50 POINTS, AND I'LL MARK BRAINLIEST!

9966 [12]

Hello!

I believe the most logical answer is 7⋅10^9

I hope it helps!

8 0

3 years ago