END OF UNIT MATH TEST PLEASE HELP

What is (3/4)³ in fraction form

suter [353]

27/64 . because you do 3/4 to the third power (3x) which gives you that

6 0

2 years ago

Please help solve the inequality.

gulaghasi [49]

Answer:

Are there two you're asking for?

#2: x ≥ 9

#3: x > -2

Step-by-step explanation:

#2 Work:

x - 13 ≥ -4

Step 1: Add 13 to both sides

x - 13 + 13 ≥ -4 + 13

Add -4 and 13 to get 9.

x ≥ 9

#3 Work:

x/2 > -1

Multiply both sides by 2. Since 2 is positive, the inequality direction remains the same

x/2 * 2 > -1 * 2

x > -2

5 0

2 years ago

If my quadrilateral has one pair of parallel sides, and one pair of equal sides, what MUST it be, and what COULD it be? (18 Poin

Zepler [3.9K]

It could be a rectangle

8 0

2 years ago

Given log Subscript 3 Baseline 2 almost-equals 0. 631 and log Subscript 3 Baseline 7 almost-equals 1. 771, what is log Subscript

kari74 [83]

You can use the properties of logarithm to get to the solution.

The approximate value for given term is given by

$log_3(14) \approx 2.402$

<h3>What is logarithm and some of its useful properties?</h3>

When you raise a number with an exponent, there comes a result.

Lets say you get

$a^b = c$

Then, you can write 'b' in terms of 'a' and 'c' using logarithm as follows

$b = log_a(c)$

Some properties of logarithm are:

$log_a(b) = log_a(c) \implies b = c\\\\\log_a(b) + log_a(c) = log_a(b \times c)\\\\log_a(b) - log_a(c) = log_a(\frac{b}{c})$

<h3>Using the above properties</h3>

$log_3(2) + log_3(7) = log_3(2 \times 7) = log_3(14)\\\\0.631 + 1.771 = log_3(14)\\\\log_3(14) = 2.402$

Thus,

The approximate value for given term is given by

$log_3(14) \approx 2.402$

Learn more about logarithm here:

brainly.com/question/20835449

5 0

1 year ago

Read 2 more answers

The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

i)

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0

2 years ago