What is 25 divided by 2350

Can they be written as a fraction and if so why

Kruka [31]

The first one can be written as a fraction because it is a repeating decimal.

The second one can be written as a fraction because it is a repeating decimal.

The third one can not be written as a fraction since it is nether terminating nor repeating.

The last one can be written as a fraction since it is a terminating decimal.

4 0

3 years ago

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URGENT pleaseee help 10 points and brainlist if right!!!!<br> Use the Sohcahtoa method!!

Vsevolod [243]

Answer: x is 36 degrees

Step-by-step explanation:

To solve this triangle, we can use sin -1 x = 14 / 24. This gives us 35.69 degrees, but in your answer choices you only get integers. We can round this up to 36 degrees to get the answer.

8 0

2 years ago

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In the United States, voters who are neither Democrat nor Republican are called Independent. It is believed that 11% of voters a

Radda [10]

Answer:

a) 0.0214 = 2.14% probability that none of the people are Independent.

b) 0.8516 = 85.16% probability that fewer than 6 are Independent.

c) 0.8914 = 89.14% probability that more than 2 people are Independent.

Step-by-step explanation:

For each people, there are only two possible outcomes. Either they are independent, or they are not. For each person asked, the probability of them being Independent voters is the same. This means that 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.

It is believed that 11% of voters are Independent.

This means that $p = 0.11$

A survey asked 33 people to identify themselves as Democrat, Republican, or Independent.

This means that $n = 33$

A. What is the probability that none of the people are Independent?

This is P(X = 0). So

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

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

0.0214 = 2.14% probability that none of the people are Independent.

B. What is the probability that fewer than 6 are Independent?

This is

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

So

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

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

$P(X = 1) = C_{33,1}.(0.11)^{1}.(0.89)^{32} = 0.0872$

$P(X = 2) = C_{33,2}.(0.11)^{2}.(0.89)^{31} = 0.1724$

$P(X = 3) = C_{33,3}.(0.11)^{3}.(0.89)^{30} = 0.2202$

$P(X = 4) = C_{33,4}.(0.11)^{4}.(0.89)^{29} = 0.2041$

$P(X = 5) = C_{33,5}.(0.11)^{5}.(0.89)^{28} = 0.1463$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0214 + 0.0872 + 0.1724 + 0.2202 + 0.2041 + 0.1463 = 0.8516$

0.8516 = 85.16% probability that fewer than 6 are Independent.

C. What is the probability that more than 2 people are Independent?

This is:

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

In which

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

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

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

$P(X = 1) = C_{33,1}.(0.11)^{1}.(0.89)^{32} = 0.0872$

$P(X < 2) = 0.0214 + 0.0872 = 0.1086$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1086 = 0.8914$

0.8914 = 89.14% probability that more than 2 people are Independent.

8 0

2 years ago

Can anyone help me with this?

erma4kov [3.2K]

Answer: 16y² - x²

Step-by-step explanation: The - sign means a difference, so the choices with + signs are eliminated (though 64x² and 9 are squares)

10 is not a square so that one is eliminated (though the y² and the 4x² are squares)

16 is the square of 4, y² is the square of y, and x² is the square. That expression shows a difference of squares.

3 0

3 years ago

Writing an Equa

kakasveta [241]

Answer:

Step-by-step explanation:

(3,1) lies below the vertex, so the parabola opens downwards.

Equation for a down-opening parabola with vertex (2,3):

y = a(x-2)² + 3, a<0

Plug in (3,1) and solve for a:

1 = a1² + 3

a = -2

y = -2(x-2)² + 3

6 0

2 years ago