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3 years ago

What is two decimal that are equivalent to 2.145

Mathematics

2 answers:

Amanda [17]3 years ago

5 0

Answer:

Two decimals that are equivalent to 2.145 are 2.1450 and 2.145.

Step-by-step explanation: If you are trying to find an equivalent decimal to anything the easiest way is to add zeros to the end.

kolezko [41]3 years ago

5 0

Answer:

2.145000 and 2.14500

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2 years ago

Three years ago, Sherman was twice as old as Clive. If Clive’s present age is c, which expression represents Sherman’s present a

OLEGan [10]

First establish variables for current age.

Clive's current age = c
Sherman's current age = s

"Three years ago,"
Clive was: (c - 3)
Sherman was: (s - 3)

"Sherman was twice as old as Clive."
(s - 3) = 2(c - 3)

Then in terms of Sherman's present age, distribute 2, combine like terms
s - 3 = 2c - 6
s = 2c - 6 + 3
s = 2c - 3

7 0

3 years ago

About 3% of children in the United States are allergic to peanuts. Choose three children at random and let the random variable X

Charra [1.4K]

Answer:

0.029991 = 2.9991% conditional probability that exactly two of the children will be allergic to peanuts, given that at least one of the three children suffers from this allergy

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are allergic to peanuts, or they are not. The probability of a children being allergic to peanut butter is independent of other children. 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.

3% of children in the United States are allergic to peanuts.

So $p = 0.03$

Three children:

So n = 3,

The probabilities are:

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

$P(X = 0) = C_{3,0}.(0.03)^{0}.(0.97)^{3} = 0.912673$

$P(X = 1) = C_{3,1}.(0.03)^{1}.(0.97)^{2} = 0.084681$

$P(X = 2) = C_{3,2}.(0.03)^{2}.(0.97)^{1} = 0.002619$

$P(X = 3) = C_{3,3}.(0.03)^{3}.(0.97)^{0} = 0.000027$

What is the conditional probability that exactly two of the children will be allergic to peanuts, given that at least one of the three children suffers from this allergy?

This is

$p = \frac{P(X = 2)}{P(X \geq 1)}$

In which

$P(X \geq 1) = 0.084681 + 0.002619 + 0.000027 = 0.087327$

$P(X = 2) = 0.002619$

$p = \frac{0.002619}{0.087327} = 0.03$

0.029991 = 2.9991% conditional probability that exactly two of the children will be allergic to peanuts, given that at least one of the three children suffers from this allergy

8 0

3 years ago