If you are able to save, $150 dollars per month-how many months will it take you to

Mathematics

2 answers:

Katen [24]3 years ago

6 0

Answer:

16.66months

Step-by-step explanation:

2500/150

Nitella [24]3 years ago

4 0

I agree with the person above 16.66

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Match the part of the circle with the term that best describes it.

nordsb [41]

GH: chord
M: point of tangency
MJ: diameter
J: center
MH: radius
GH: secant

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2 years ago

Factor completely 3a4y3 − 12a3y2 + 6a2y.

Firlakuza [10]

Step-by-step explanation:

$3a^4y^3-12a^3y^2+6a^2y=(3a^2y)(a^2y^2)-(3a^2y)(4ay)+(3a^2y)(2)\\\\=(3a^2y)(a^2y^2-4ay+2)$

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I am confused on how to do this

babymother [125]

So, u would start by expanding the bracket and ignoring the 2 at the moment.
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3 years ago

12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests

Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

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 combinatios 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.

In this problem, we have that:

12% of children are nearsighted. This means that $p = 0.12$ .

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so $n = 18$