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3 years ago

Use the distributive property to rewrite each expression in it's equivalent form 9a-3/3 ,Just 3 please

Mathematics

1 answer:

pentagon [3]3 years ago

8 0

Hahahah i think i already did this for you buttttt

the denominator can be distributed to both parts of 9a and -3

so you can rewrite it as (9a/3) and (-3/3)

to simplify (9a/3) is equal to 3a and (-3/3) is equal to -1

so the answer is 3a-1

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What theorem or postulate can be used to justify that the two triangles are congruent? Write the congruence

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Answer:

Step-by-step explanation:

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2 years ago

What is the true solution to the equation below? In e^in x + in e^in x^2 = 2 in 8

mariarad [96]

To solve the given equation, w need to review some rules:
$(1) \ ln \ e = 1 \\ 
(2) \ ln \ b^{a} = a*ln \ b \\
(3) \ ln \ a + ln \ b = ln \ (ab) \\
(4) \ ln \ a - ln \ b = ln \ \frac{a}{b} \\$

The given equation is :
$ln \ e^{ln \ x} + ln \ e^{ln \ x^{2}} = 2 \ ln \ 8$
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removing the nature logarithm from both sides

$x^{3} = 64 = 4^{3}$
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So, the correct answer is option (2) ⇒⇒⇒⇒ x = 4

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2 years ago

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Fudgin [204]

Answer:

D. 8

Step-by-step explanation:

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2 years ago

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Can someone help with these 3?

adell [148]

1) 5 +8i
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3 years ago

Isle Royale, an island in Lake Superior, has provided an important study site of wolves and their prey. Of special interest is t

STatiana [176]

Answer:

Part a) Probability that a moose in that age group is killed by a wolf

P (0.5 year) = 0.361

P (1 - 5) = 0.159

P (6 - 10) = 0.267

P (11 - 15) = 0.203

P (16 - 20) = 0.010

Part b)

Expected age of a moose killed by a wold

μ = 5.61 years

Stantard deviation of the ages

σ = 4.97 years

Explanation:

1) Start by arranging the table to interpret the information:

Age of Moose in years Number killed by woolves

Calf (0.5) 107

1 - 5 47

6 - 10 79

11 - 15 60

16 - 20 3

You can now verify the total number of moose deaths identified as wolf kills: 107 + 47 + 79 + 60 + 3 = 296, such as stated in the first part of the question.

2) First question: a) For each age, group, compute the probability that a moos in that age group is killed by a wolf.

i) Formula:

Probability = number of positive outcomes / total number of events.

ii) Probability that a moose in an age group is killed by a wolf = number of moose killed by a wolf in that age group / total number of moose deaths identified as wolf kills.

iii) P (0.5 year) = 107 / 296 = 0.361

iv) P (1 - 5) = 47 / 296 = 0.159

v) P (6 - 10) = 79 / 296 = 0.267

vi) P (11 - 15) = 60 / 296 = 0.203

vii) P (16 - 20) = 3 / 296 = 0.010

3) Second question: b) Consider all ages in a class equal to the class midpoint. Find the expected age of a moose killed by a wolf and the standard deviation of the ages.

i) Class midpoint

0.5 0.5

1 - 5 (1 + 5) / 2 = 3

6 - 10 (6 + 10) / 2 = 8

11 - 15 (11 + 15) / 2 = 13

16 - 20 (16 + 20) = 18

ii) Expected age of a moose killed by a wolf = mean of the distribution = μ

μ = Sum of the products of each probability times its age (mid point)

μ = 0.5 ( 0.361) + 3 ( 0.159) + 8 ( 0.267) + 13 ( 0.203) + 18 ( 0.010) = 5.61 years

μ = 5.61 years ← answer

iii) Stantard deviation of the ages = σ

σ = square root of the variace

variance = s = sum of the products of the squares of the differences between the mean and the class midpoint time the probability.

s = (0.5 - 5.61)² (0.361) + (3 - 5.61)² ( 0.159) + (8 - 5.61)² ( 0.267) + (13 - 5.61)² ( 0.203) + (18 - 5.61)² ( 0.010) = 24.65

σ = √ (24.65) = 4.97 years ← answer

7 0

2 years ago