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2 years ago

Solve the proportion 8/3=g/3 ... A.1/8 B.9/8 C.8 OR D.8/9

Mathematics

1 answer:

andrey2020 [161]2 years ago

4 0

Answer:

g = 8. (Answer C)

Step-by-step explanation:

We are given 8/3 = g/3.

If we multiply both sides by 3, this will eliminate the fractions, and we will be left with g = 8. (Answer C).

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What’s 127.2 divided by 14

Burka [1]

9.1 *rounded tot he nearest 10th place*

8 0

3 years ago

a certain pharmaceutical company know that, on average 4% of a certain type of pill has an ingredient that is below the minimum

FinnZ [79.3K]

Using the binomial distribution, it is found that there is a 0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

For each pill, there are only two possible outcomes, either it is acceptable, or it is not. The probability of a pill being acceptable is independent of any other pill, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

The sample has 20 pills, hence $n = 20$ .

100 - 4 = 96% are acceptable, hence $p = 0.96$

The probability that fewer that 5 in a sample of 20 pills will be acceptable is:

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

In which

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

$P(X = 0) = C_{20,0}.(0.96)^{0}.(0.04)^{20} = 0$

$P(X = 1) = C_{20,1}.(0.96)^{1}.(0.04)^{19} = 0$

$P(X = 2) = C_{20,2}.(0.96)^{2}.(0.04)^{18} = 0$

$P(X = 3) = C_{20,3}.(0.96)^{3}.(0.04)^{17} = 0$

$P(X = 4) = C_{20,4}.(0.96)^{4}.(0.04)^{16} = 0$

0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

A similar problem is given at brainly.com/question/24863377

8 0

2 years ago

Please help with these two math problems 1. 7^x ÷ 7^y = 2. z^10x^y ÷ z^5 =

IceJOKER [234]

Answer:

1. 7^(x-y)

2. z^5x^y

Step-by-step explanation:

The applicable rule of exponents is ...

(a^b)/(a^c) = a^(b-c)

$\dfrac{7^x}{7^y}=\boxed{7^{x-y}}$

$\dfrac{z^{10}x^y}{z^5}=x^yz^{10-5}=\boxed{z^5x^y}$

8 0

3 years ago

An insurance company issues life insurance policies in three separate categories: standard,preferred,and ultra- preferred. Of th

slega [8]

Answer:

a) 0.0002

b) 0.0057

c) 0.0364

Step-by-step explanation:

Lets start by stating the probabilities of a person belonging to each policy:

Standard: 0.3

Preferred: 0.5

Ultra- Preferred: 0.2

The probability of person belonging to each policy AND dying in the next year:

Standard: 0.3 x 0.015 = 0.0045

Preferred: 0.5 x 0.002 = 0.001

Ultra- Preferred: 0.2 x 0.001 = 0.0002

a) The probability a ultra - preferred policy holder dies in the next year is 0.001. To find the probability of a person being both a ultra - preferred policy holder AND die in the next year is: 0.001 x 0.2= 0.0002

b) The probability is given by adding the probabilities calculated before :

0.0045 + 0.001 + 0.0002 = 0.0057

c) We use the results above again. This is 0.0002 / (0.001 + 0.0045). The answer comes out to be 0.0364

6 0

3 years ago

Evaluate 7+5^2*4 Algebra 1

vagabundo [1.1K]

Hi there!

Answer:

=107

*The answer must have a positive sign.*

Step-by-step explanation:

<h2>Lesson: Order of operations</h2>

Parenthesis

Exponents

Multiply

Divide

Add

Subtract

It can also go left to right.

First, you do exponents.

$5^2=5*5=25$

$7+25*4$

Then, you multiply from left to right.

$25*4=100$

Finally, you add from left to right.

$100+7=107$

I hope this helps you!

Have a nice day! :)

-Charlie

8 0

3 years ago

Read 2 more answers