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4 years ago

What is -1/2 subtract -5/6

Mathematics

2 answers:

earnstyle [38]4 years ago

5 0

Answer: .33

Step-by-step explanation: Division can be frustrating with fractions, that's why i like to translate them into decimals and then solve them on a piece of paper

Vinvika [58]4 years ago

5 0

Answer:

1/3

Step-by-step explanation:

-0.5 - (-0.83333333333)

-0.5 + 0.83333333333

0.333333333333

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An oil drilling company ventures into various locations, and its success or failure is independent from one location to another.

densk [106]

Answer:

a) 18.77% probability that the driller drills at 10 locations and has 1 success

b) 75.60% probability that the driller drills at 10 locations and has at least 2 success

Step-by-step explanation:

For each drill, there are only two possible outcomes. Either it is a success, or it is not. The probability of a drill being a success is independent of other drills. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

10 locations

This means that $n = 10$

Suppose the probability of a success at any specific location is 0.25.

This means that $p = 0.25$

(a) What is the probability that the driller drills at 10 locations and has 1 success?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{10,1}.(0.25)^{1}.(0.75)^{9} = 0.1877$

18.77% probability that the driller drills at 10 locations and has 1 success

(b) What is the probability that the driller drills at 10 locations and has at least 2 success?

Either there are less than 2 success, or there are at least 2. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

We want $P(X \geq 2)$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.25)^{0}.(0.75)^{10} = 0.0563$

$P(X = 1) = C_{10,1}.(0.25)^{1}.(0.75)^{9} = 0.1877$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0563 + 0.1877 = 0.2440$

$P(X < 2) = P(X = 0) + P(X = 1) = 1 - 0.244 = 0.756$

75.60% probability that the driller drills at 10 locations and has at least 2 success

8 0

3 years ago

Constance is saving money to buy a new bicycle that costs 195.75.She already 40 saved and plans to save 8 each week. How many we

Anna11 [10]

Answer:

19.5 weeks

Step-by-step explanation:

8x19.5=156+40=196

8 0

3 years ago

Does anyone know the answer to these

vladimir1956 [14]

53) The metric unit of mass is kilogram. It is the mass of a piece of platinum alloy that keep in a museum in Paris. One kg (kilogram) is equal to 1000 g (gram). One kg (mass) weighs 2.205 pounds on the surface of the Earth.

54) The unit used to measure weight in the metric system is the gram. Other metric units for weight include the kilogram, the metric ton and the milligram

55) not sure about this one

56)

6 0

3 years ago

Factorise x^12 y^4 - x^4 y^12<br><br>PLEASE PLEASE explain in step by step Pleasseeeeee

QveST [7]

Answer:

Step-by-step explanation:

First, look at your terms. You have two terms and two variables in each. So they can be factored out because you can distribute them back in easily.

Focus one variable first, x. You cant factor an x^12 out of an x^4. You can, but you really don't want to, it'll turn your problem into a mess. So the only thing you can do is factor out the lower exponent, x^4.

$x^4(x^8y^4-y^1^2)$

We arent done yet because the terms still share the y variable. You cant factor a y^12 from a y^4. This is the same problem with the x variable. So take out the lower one, y^4.

$x^4y^4(x^8-y^8)$

8 0

3 years ago

Help pls i will give you brainlist but make sure its correct help ASAP

Strike441 [17]

It is 1,6 dollars for each pendant :)

6 0

3 years ago