If two circles have the same circumference what do you know about their areas

PWEASE HELP I WILL MARK CHU BRAINLIST >:3

Vanyuwa [196]

Answer:

The scale factor is

Is a reduction

Step-by-step explanation:

we know that

The scale factor is the ratio of the corresponding side of resulting triangle by the corresponding side of the first triangle

Let

z-----> the scale factor

x------> the corresponding side of resulting triangle

y------> the corresponding side of the first triangle

so

In this problem we have

substitute

The scale factor is less than one

therefore

Is a reduction

6 0

3 years ago

If X denotes the number of heads in n tosses of a coin, what is the standard deviation of the random variable X? Does this stand

vfiekz [6]

Answer:

$\sigma=\sqrt{np(1-p)}=\sqrt{n*0.5(1-0.5)}=\frac{\sqrt{n}}{2}$

As we can see the deviation is proportional to the value of n and if n increase then the deviation increases too. So then the deviation would be larger when n gets larger.

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".

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:

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

Solution to the problem

We can define the following random variable X ="Number of heads in n tosses of a coin".

We assume that the coin is fair and then $P(H)= 0.5$ for any trial so then we can model X with the following distribution:

$X \sim Bin (n, p=0.5)$

For this distribution the mean and variance are given by:

$E(X)=np=n*0.5=\frac{n}{2}$

$Var(X)= np(1-p) = n*0.5*(1-0.5) = 0.25 n = \frac{n}{4}$

And the deviation would be just the square root of the variance and we got:

$\sigma=\sqrt{np(1-p)}=\sqrt{n*0.5(1-0.5)}=\frac{\sqrt{n}}{2}$

Does this standard deviation get larger or smaller when n gets larger?

As we can see the deviation is proportional to the value of n and if n increase then the deviation increases too. So then the deviation would be larger when n gets larger.

3 0

3 years ago

On a map, 1/2 inch represents 10 actual miles. Two towns that are 4 1/2 inches apart on this map are how many actual miles apart

dusya [7]

The answer is E.90

1/2 inches = 10 miles
9(1/2) inches = 4 1/2 inches
9(10) miles = 90 miles

7 0

3 years ago

(10 points)Assume IQs of adults in a certain country are normally distributed with mean 100 and SD 15. Suppose a president, vice

vesna_86 [32]

Answer:

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Step-by-step explanation:

To solve this question, we need to use the binomial and the normal probability distributions.

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.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability the president will have an IQ of at least 107.5

IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that $\mu = 100, \sigma = 15$

This probability is 1 subtracted by the p-value of Z when X = 107.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{107.5 - 100}{15}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 probability that the president will have an IQ of at least 107.5.

Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{130 - 100}{15}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:

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

In which

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

$P(X = 0) = C_{2,0}.(0.9772)^{2}.(0.0228)^{0} = 0.9549$

$P(X \geq 1) = 1 - P(X = 0) = 0.0451$

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

What is the probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130?

0.3085 probability that the president will have an IQ of at least 107.5.

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Independent events, so we multiply the probabilities.

0.3082*0.0451 = 0.0139

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

8 0

3 years ago

So I need help finding the cross section of all the shapes, so any cross section experts mind helping?

Kisachek [45]

Answer:

Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao

Step-by-step explanation:

3 0

3 years ago