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3 years ago

What fraction is equivalent to a repeating decimal

Mathematics

1 answer:

klasskru [66]3 years ago

6 0

Answer:

0.3333. Equivalent to 1/3

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Write an equation and solve:

11Alexandr11 [23.1K]

Answer

Find out how many months will it take for Jeff to weigh the same as Darius .

To prove

As given

Jeff and Darius are trying to gain weight for the football team.

Jeff weighs 120 pounds and is gaining ten pounds per month.

Darius weighs 150 pounds and is gaining 4 pounds per month.

m = represented the number of months will taken by the Jeff to weigh the same as Darius .

Than the equation become in the form

120 + 10m = 150 +4m

150 - 120 = 10m - 4m

30 = 6m

$m = \frac{30}{6}$

m = 5 months

Therefore in 5 months will it take for Jeff to weigh the same as Darius.

4 0

4 years ago

a quadrilateral has angles measuring 105 degrees, 80 degrees, and 45 degrees what is the measure of the 4th angle

Bond [772]

Answer: 130 degrees

Step-by-step explanation:

105+80+45+x=360

230+x=360

-230 both sides

X=130 degrees

6 0

3 years ago

Plsssss helppppp i need to do this work in 10 min plss helpp

Over [174]

Answer:

3/2 or 1.5

Step-by-step explanation:

Change in y/ Change in x

3/2

1.5

You're welcome and don't forget to click THANKS

7 0

3 years ago

Read 2 more answers

Sketch the graph of each line

sveta [45]

Use Desmos it helps you with questions like that

3 0

3 years ago

An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed

masya89 [10]

Answer:

$P(X\leq 1) = 0.331$

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

$P(X = x) = \frac{y^xe^{-y}}{x!}$

Where y represents λ (average)

y = 2.3

Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment

Using probability notations;

$P(X\leq 1) = P(X=0) + P(X =1)$

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

$P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}$

$P(X = 0) = \frac{1 * e^{-2.3}}{1}$

$P(X = 0) = e^{-2.3}$

$P(X = 0) = 0.10025884372$

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

$P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}$

$P(X = 1) = \frac{2.3 * e^{-2.3}}{1}$

$P(X = 1) =2.3 * e^{-2.3}$

$P(X = 1) = 2.3 * 0.10025884372$

$P(X = 1) = 0.23059534055$

$P(X\leq 1) = P(X=0) + P(X =1)$

$P(X\leq 1) = 0.10025884372 + 0.23059534055$

$P(X\leq 1) = 0.33085418427$

$P(X\leq 1) = 0.331$

Hence, the required probability is 0.331

6 0

3 years ago