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3 years ago

A true-false test has 12 questions. What is the probability of guessing the correct answers to all of the questions?

Mathematics

2 answers:

alexandr1967 [171]3 years ago

7 0

Each t or f has 2 possibilities so 2^12 or 4096

AleksAgata [21]3 years ago

7 0

1/2 is the answer. There are two choices, True and False so you have a 50/50 chance of being right

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Which expression is equivalent to 42 + 30? (1 point)

ikadub [295]

Well lets see A, B and D are too large in the parenthesis that the numbers outside of the parenthesis will use to multiply considering 42 + 30.
6(7 + 5)
6(12)
72

6 0

3 years ago

Read 2 more answers

The type of equation given below is in point-slope form <br> y=1/2x-5 true or false

kykrilka [37]

Answer:

false

Step-by-step explanation:

point slope form : y - y1 = m( x - x1 )

the equation y = 1/2x - 5 does not follow this therefore the answer is false

the equation y = -1/2x - 5 is instead put in slope intercept form , y = mx + b

8 0

2 years ago

Richard's cumulative GPA for 3 semesters_ was 2.0 for 42 credits. His fourth semester GPA was 4.0 for 14 course units. What is R

Phoenix [80]

Answer:

Richard's cumulative GPA = 2.5

Step-by-step explanation:

Richard's cumulative GPA for 3 semesters was 2.0 for 42 credits.

Average = 2

Total credits = 42

Average = sum of terms/total no of terms

2 = sum of terms/42

sum of terms = 84 ....(1)

His fourth semester GPA was 4.0 for 14 course units

Average = 4

Total credits = 14

sum of terms = 56 .....(2)

Average for 4 semesters,

$A=\dfrac{84+56}{42+14}\\\\A=2.5$

Hence, Richard's cumulative GPA for all 4 semesters is 2.5.

8 0

3 years ago

\lim _{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)

Vinvika [58]

$\displaystyle \lim_{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)$

Both the numerator and denominator approach 0, so this is a candidate for applying L'Hopital's rule. Doing so gives

$\displaystyle \lim_{x\to 0}\left(2\ln(1+3x)+\dfrac{6x}{1+3x}+\cos(x)\tan(3x)+3\sin(x)\sec^2(x)-6x^2}{3\sin(3x)}\right)$

This again gives an indeterminate form 0/0, but no need to use L'Hopital's rule again just yet. Split up the limit as

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} + \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} \\\\ + \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} + \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} \\\\ - \lim_{x\to0}\frac{6x^2}{3\sin(3x)}$

Now recall two well-known limits:

$\displaystyle \lim_{x\to0}\frac{\sin(ax)}{ax}=1\text{ if }a\neq0 \\\\ \lim_{x\to0}\frac{\ln(1+ax)}{ax}=1\text{ if }a\neq0$

Compute each remaining limit:

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{\ln(1+3x)}{3x} \times \lim_{x\to0}\frac{3x}{\sin(3x)} = \frac23$

$\displaystyle \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\frac{1}{1+3x} = \frac23$

$\displaystyle \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\cos(x)}{\cos(3x)} = \frac13$

$\displaystyle \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\sin(x)}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\sec^2(x) = \frac13$

$\displaystyle \lim_{x\to0}\frac{6x^2}{3\sin(3x)} = \frac23 \times \lim_{x\to0}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}x = 0$

So, the original limit has a value of

2/3 + 2/3 + 1/3 + 1/3 - 0 = 2

6 0

3 years ago

What is the area of the square? Perimeter =108mi. 27 width

Leni [432]

Answer:

729 square units

Step-by-step explanation:

If the width of the square is 27, then the length is also 27. After all, this is a square with four equal sides.

The area of a square is A = s^2, where s is the length of one side.

Here, A = 27^2 = 729 square units.

4 0

4 years ago