Which will have a negative number?

Marci has45 red buttns and 60 yellow butto a she wants to use all of the butt s a d divide each color equally into boxes what is

aksik [14]

7 because 45+60=105÷7=15

5 0

3 years ago

Solve: x^2 = 81 x^2 = -9 x^2 = 20 3x^2 = 125

sergeinik [125]

x^2 = 81
x = ±√81
x = ±9

x^2 = -9 no solutions

x^2 = 20
x = ±√20
x = ±2√5

 $3x^2 = 125 \\ 
x^2 = \frac{125}{3} \\ x=б \sqrt{ \frac{125}{3} }$

3 0

3 years ago

Software filters rely heavily on ""blacklists"" (lists of known ""phishing"" URLs) to detect fraudulent e-mails. But such filter

crimeas [40]

Answer:

a) $X \sim Binom(n=16, p=0.2)$

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:

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

The expected value is given by this formula:

$E(X) = np=16*0.2=3.2$

b) $P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815$

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".

Part a

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

$X \sim Binom(n=16, p=0.2)$

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:

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

The expected value is given by this formula:

$E(X) = np=16*0.2=3.2$

Part b

For this case we want this probability:

$P(X=0)$

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:

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

And using this function we got:

$P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815$

6 0

3 years ago

Can someone solve this, you don’t have to type an explanation, thank you.

iogann1982 [59]

Answer:

70

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Suppose that a researcher is designing a survey to estimate the proportion of adults in your state who oppose a proposed law tha

irinina [24]

Answer:

$n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401$

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Solution to the problem

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

The margin of error desired for this case is $ME= \pm 0.02$ equivalent to 2% points

For this case we need to assume a confidence level, let's assume 95%. And since we don't have prior estimation for the population proportion of interest the best value to do an approximation is $\hat p =0.5$

In order to find the critical value we need to take in count that we are finding the margin of error for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.96$

Now we have all the values needed and if we replace into equation (b) we got:

$n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401$

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.

5 0

3 years ago