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2 years ago

Can you please help me with my math thank you

Mathematics

1 answer:

gayaneshka [121]2 years ago

7 0

2•8.2•5.1=83.64 square feet.
ANSWER: 83.64 Square Feet

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What is the solution set for the given inequality if the replacement set for r is {5, 6, 7, 8, 9, 10}? 2r ≤ 3r – 8 A. {8, 9, 10}

TiliK225 [7]

$2r\leq3r-8\\
r\geq8\\\Downarrow\\
\boxed{\{8,9,10\}}$

8 0

3 years ago

Express 112 as a product of prime numbers in index form?

alexgriva [62]

112
/ \
2 56
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2 28
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So, the answer is 2 x 2 x 2 x 7 or 2^3 x 7

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3 years ago

Find the slope of the line passing through the given points.

deff fn [24]

Answer:

<h3>The answer is - 3</h3>

Step-by-step explanation:

The slope of a line given two points can be found by using the formula

$m = \frac{ y_2 - y _ 1}{x_ 2 - x_ 1} \\$

From the question we have

$m = \frac{8 - 5}{ - 8 - - 7} = \frac{3}{ - 8 +7 } = - \frac{3}{1} = - 3 \\$

We have the final answer as

Hope this helps you

5 0

2 years ago

Elsa and Anna share $2000 in the ratio 5 : 3. How much does each of them get?

Karolina [17]

Answer:

Let them be "x"

5x+3x= $2000

8x= $2000

x= 2000/8

x= 250

5* 250= 1250.

3*250= 750.

5 0

2 years ago

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

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