All categories

2 years ago

I don’t know how to answer the questions

Mathematics

1 answer:

ololo11 [35]2 years ago

7 0

Answer:

all work is pictured and shown

You might be interested in

A cow can give 5 gallons of milk if it eats 2 pounds of grass. Express x in terms of y, where y is the number of gallons of milk

spayn [35]

Answer:

x = 2.5y

Step-by-step explanation:

If a cow can give 5 gallons of milk after eating 2 pounds of grass.

Let's consider 'x' is the number of pounds of grass it eats and 'y' is the number of gallons of milk the cow gives.

Hence,

5y = 2x

We will simplify it to get what x equal to y by dividing 2 on both side we find,

x = 2.5y

x is equal to 2.5 y

By eating 1 pound of grass cow can give 2.5 gallons of milk.

7 0

3 years ago

Work out 4 1/2 - 1 1/3

Ainat [17]

Answer: 19/6

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

What is seven over twenty one take away one over twenty one in its simplest form

natulia [17]

It would be 6/21 but simplified it would be 2/7

5 0

3 years ago

Can someone Help me in this problem?

Fed [463]

If you have to write the equation then it would be "5k + 6.7"

5 0

3 years ago

A sequence of Bernoulli trials consists of choosing components at random from a batch of components. A selected component is eit

Drupady [299]

Answer:

a) 0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.

Step-by-step explanation:

A sequence of Bernoulli trials composes the binomial 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 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 probability that a selected component is non-defective is 0.8

This means that $p = 0.8$

a) Three non-defective components in a batch of seven components.

This is P(X = 3) when n = 7. So

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

$P(X = 3) = C_{7,3}.(0.8)^{3}.(0.2)^{4} = 0.0287$

0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 8 non-defective components are drawn before the first defective component is chosen.

Now the order is important, so the we just multiply the probabilities.

8 non-defective, each with probability 0.8, and then a defective, with probability 0.2. So

$p = (0.8)^8*0.2 = 0.0336$

0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.

4 0

2 years ago