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1 year ago

Confidence intervals for the means provide an estimate for where the true mean lies.

Mathematics

1 answer:

stepladder [879]1 year ago

3 0

It is true that the confidence intervals for the mean provide an estimate for where the true mean lies.

In statistics, a confidence interval denotes the likelihood that a population parameter will fall between a set of values for a given proportion of the time. A confidence interval depicts the likelihood that a parameter will fall between two values near the mean. Confidence intervals quantify the degree of uncertainty or certainty in a sampling procedure.

The mean is a basic mathematical average of two or more values. There are two sorts of means that may be calculated: the arithmetic mean and the geometric mean. A mean tells you the average of a bunch of values, which helps you contextualize each data point.

To learn more about confidence interval, visit :

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The volume of a cylinder is 8177 cm³. If the radius is 3 cm, what is the height

Virty [35]

Answer:

None of the options are correct

О E. 289.2 cm

Step-by-step explanation:

Volume of a cylinder = v = л r² h

л = 3.1416 (aprox)

r = radius

h = height

8177 = 3.1416 * 3² * h

8177 = 3.1416 * 9 * h

8177 = 28.2744 * h

8177/282744 = h

h = 289.2 cm

4 0

2 years ago

At the bakery where you work, loaves of bread are supposed to weigh 1 pound, with a standard deviation of 0.13 pounds. You belie

saveliy_v [14]

Answer:

Kindly check explanation

Step-by-step explanation:

Given that:

H0 : μ = 1

H1 : μ > 1

Sample size, n = 20 ; xbar = 1.05; Standard deviation, s = 0.13

Test statistic;

t because sample size is < 30

(xbar - μ) ÷ (s/sqrt(n))

(1.05 - 1) ÷ (0.13 / sqrt(20))

0.05 ÷ 0.0290688

= 1.7200572

Using a p value calculator :

With alpha = 0.05 ; df = n - 1 = 20 - 1 = 19

P value = 0.05084

P value > alpha

We fail to reject H0

There is not sufficient evidence to support that the new personnel are producing loaves that are heavier than 1 pound

8 0

3 years ago

You are at a stall at a fair where you have to throw a ball at a target. There are two versions of the game. In the first

Tomtit [17]

Answer:

$P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729$

And the probability of loss with the first wersion is 0.729

$P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774$

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Alternative 1

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=3, p=0.1)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

We can find the probability of loss like this P(X=0) and if we find this probability we got this:

$P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729$

And the probability of loss with the first wersion is 0.729

Alternative 2

Let Y the random variable of interest, on this case we now that:

$Y \sim Binom(n=5, p=0.05)$

The probability mass function for the Binomial distribution is given as:

$P(Y)=(nCy)(p)^y (1-p)^{n-y}$

Where (nCx) means combinatory and it's given by this formula:

$nCy=\frac{n!}{(n-y)! y!}$

We can find the probability of loss like this P(Y=0) and if we find this probability we got this:

$P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774$

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

4 0

4 years ago

Find the ratio of the corresponding sides of triangle a to triangle<br> b.

saul85 [17]

Ratio of corresponding sides of triangle a to triangle b is a/b

8 0

3 years ago

Can someone check questions?

Soloha48 [4]

1
The first one is correct.

2
A hexagon has the weird property of having the radius and the side being of exactly the same length

The formula is A = 3 * sqrt(3) * a^2 / 2
where a = the side or radius
A = 3 * sqrt(3) * 5^2 / 2
A = 3 * sqrt(3) * 25 / 2
A = 1.5 * sqrt(3) * 25
A = 64.95

A <<<< Answer (the first one is the answer.)

C
This one seems so much more entailed. Do you know what the cos law is? That's what I will use to find c.

c^2 = a^2 + b^2 - 2*ab* cos(C)
C = 33 degrees
a = 2.75 miles
b = 1.32 miles

c^2 = 2.75^2 + 1.32^2 - 2 * 2.75 * 1.32 * cos(33)
c^2 = 9.3049 - 6.0887
c^2 = 3.2162
c = sqrt(3.2162) = 1.7934

Now you have to use this result to get the area. You have to use Heron's formula.

A = sqrt(s * (s - a) * (s - b) * (s - c) )
s = 1/2 the perimeter
s = 1/2 (2.75 + 1.32 + 1.7934)
s = 1/2(5.8634)
s = 2.9317
I'll finish this in the comments. I have to leave for a bit.

6 0

4 years ago