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tankabanditka [31]
3 years ago
8

What is 51% of 25.5?​

Mathematics
2 answers:
marshall27 [118]3 years ago
3 0

Answer:

The answer is 13.005

Step-by-step explanation:

Otrada [13]3 years ago
3 0

Answer:

13.005

Step-by-step explanation:

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I'LL GIVE BRAINLIEST
likoan [24]

Answer:

The answer is 300

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
2x^2+7x-15=0
Rudiy27
First step is to factorize the given expression as follows:
2x^2 + 7x - 15 = (2x-3)(x+5)
This means that the two roots of the equation are 3/2 (or 1.5) and -5

Now, we will get r and s:
we are given that the value of r is greater than the value of s, since 1.5 is greater than -5, therefore:
r = 1.5
s = -5
5 0
4 years ago
Ariana answered 36 questions correctly on her multiple choice science final and earned a grade of 45%. How many total questions
meriva

Answer:

80

Step-by-step explanation:

36 qusetions divided by .45

8 0
3 years ago
Which number is equal to 10/-3
olasank [31]
Divide 10 by -3 which equals -3.333 repeating
6 0
4 years ago
Read 2 more answers
g The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control
Rudiy27

Answer:

0.1426 = 14.26% probability that at least one of the births results in a defect.

Step-by-step explanation:

For each birth, there are only two possible outcomes. Either it results in a defect, or it does not. The probability that a birth results in a defect is independent of any other birth. This means that the binomial probability distribution is used 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.

The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC).

This means that p = \frac{1}{33}

A local hospital randomly selects five births.

This means that n = 5

What is the probability that at least one of the births results in a defect?

This is:

P(X \geq 1) = 1 - P(X = 0)

In which

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

P(X = 0) = C_{5,0}.(\frac{1}{33})^{0}.(\frac{32}{33})^{5} = 0.8574

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.8574 = 0.1426

0.1426 = 14.26% probability that at least one of the births results in a defect.

4 0
3 years ago
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