Which graph represents the solutions to the inequality |2x − 4| less than or equal to 14?

What is the answer to f(x)-3x^2-2

aleksklad [387]

Answer:

Do it yourself

Step-by-step explanation: Cheating is not a good way, what are you gonna do when you can't cheat?

6 0

3 years ago

You are collecting a sample of 60 data points from a population that you know to follow an exponential distribution. It is not a

11111nata11111 [884]

Answer:

For the exponential distribution:

$\mu = \frac{1}{\lambda}$

$\sigma^2 = \frac{1}{\lambda^2}$

We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.

$\bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = \bar X$

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}$

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

For this case we have a large sample size n =60 >30

The exponential distribution is the probability distribution that describes the time between events in a Poisson process.

For the exponential distribution:

$\mu = \frac{1}{\lambda}$

$\sigma^2 = \frac{1}{\lambda^2}$

We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.

$\bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = \bar X$

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}$

5 0

4 years ago

At the malt shop a large chocolate shake takes 8/9 of a pint of milk.If the medium shake takes 1/7 the amount of the large how m

jasenka [17]

There are several information's of immense importance in the problem. Based on these information's the answer to the question can be easily reached. The only thing that has to be kept in mind is the calculation part, otherwise it is a simple problem.
Now it is given that:
A large chocolate shake takes 8/9 of a pint of milk.
Then
Amount of milk required to make medium shake = (8/9) * (1/7) of a pint
= 8/63 of a pint
Then it can be said that for making a medium chocolate shake , 8/63 of a pint of milk is required. This is the simplest form of the fraction and it cannot be expressed as a mixed number since the denominator is higher than the numerator.

7 0

3 years ago

Solve this please! Also fill in the blanks

Annette [7]

X, 14, 2x - 16 is greater than or equal to 12, 12, add 16 to both sides, divide both sides by 2, x is greater than or equal to 14. each box filled in is represented by the commas in between each phrase. I don’t know the rest but I did get the first part.

4 0

3 years ago

1000 cylindrical metal bars of length 120 m and diameter 0.2 m , as shown below, are to be used in constructing a bridge. He bar

Troyanec [42]

Answer:

the number of of litres of paint to the nearest litre for 1000 bars is 1 litre

Step-by-step explanation:

Given that:

1000 cylindrical metal bars have :

a length of 120 m

and a diameter of 0.2 m

The volume of the cylindrical bar can be calculated by the formula:

$V= \pi r^2 h$

where ;

radius $r=\dfrac{d}{2}$

$r = \dfrac{0.2}{2}$

r = 0.1 m

the length of the cylindrical metal bars is equal to the height of the metal bar = 120 m

Thus the volume can now be calculated as follows:

$V= \pi* 0.1^2*120$

V = 3.77 m³

So; from the given information.

One litre of paint will cover 4m³

and we want to find how many litres of paint will be used for the 1000 cylindrical bars.

If 1 litre = 4m³

Then x litre = 3.77 m³

4x = 1 × 3.77

$x = \dfrac{1*3.77 }{4}$

x = 0.9425 litres

x = 1 litres to the nearest litres.

Thus ; the number of of litres of paint to the nearest litre for 1000 bars is 1 litre

4 0

4 years ago