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3 years ago

11

In taking a multiple-choice test, there are four possible answers for each problem, only one of which is correct. Assume that so

meone guesses randomly on all 25 problems.

1 answer:

Degger [83]3 years ago

7 0

Total number of events
4^25

Probabilty to get all 25 randomly,

1^(4^25)

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I need this answered in ONE minute

Helen [10]

Answer:

$x^4 + 4x^3 + 6x^2 + 7x + 2$

Step-by-step explanation:

We are asked to multiply the given polynomials.

$(x^ 2 + 3x + 1) \times (x^2 + x + 2)$

Multiply each term of the first polynomial to each term of the second polynomial.

$x^ 2 \times (x^2 + x + 2) = x^4 + x^3 + 2x^2$

$3x \times (x^2 + x + 2) = 3x^3 + 3x^2 + 6x$

$1 \times (x^2 + x + 2) = x^2 + x + 2$

Add the results

$(x^4 + x^3 + 2x^2) + (3x^3 + 3x^2 + 6x) + ( x^2 + x + 2)$

Combine the like terms

$x^4 + 4x^3 + 6x^2 + 7x + 2$

The answer is written in descending powers of x.

7 0

3 years ago

Evaluate the expression. Identify the property used in each step.

borishaifa [10]

Answer:

3/4 1/2 =3/5

Step-by-step explanation:

6 0

3 years ago

What is 3/8 equivalent to

grin007 [14]

3/8 is equivalent to 0.375.

7 0

3 years ago

GCF of 9x^2y^2 and 5x^2y^3

8090 [49]

Answer:

x^2y^2.

Step-by-step explanation:

5 0

3 years ago

D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx

sineoko [7]

Answer:

$\frac{1}{2\sqrt{5} }$

Step-by-step explanation:

Let, $\text{sin}^{-1}(\frac{x}{6})$ = y

sin(y) = $\frac{x}{6}$

$\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})$

$\frac{d}{dx}\text{sin(y)}=\frac{1}{6}$

$\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx}$ ---------(1)

$\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}$

cos(y) = $\sqrt{1-\text{sin}^{2}(y) }$

= $\sqrt{1-(\frac{x}{6})^2}$

= $\sqrt{1-(\frac{x^2}{36})}$

Therefore, from equation (1),

$\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

Or $\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

At x = 4,

$\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}$

$\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}$

$=\frac{1}{6\sqrt{\frac{36-16}{36}}}$

$=\frac{1}{6\sqrt{\frac{20}{36} }}$

$=\frac{1}{\sqrt{20}}$

$=\frac{1}{2\sqrt{5}}$

4 0

3 years ago