All categories

2 years ago

There are 120 people in a sport centre.

Mathematics

1 answer:

otez555 [7]2 years ago

6 0

Answer:

71/120

Step-by-step explanation:

You might be interested in

Find the value or measure. Assume all lines that appear to be tangent are tangent. X=

aev [14]

According to the secant-tangent theorem, we have the following expression:

$(x+3)^2=10.8(19.2+10.8)$

Now, we solve for <em>x</em>.

$\begin{gathered} x^2+6x+9=10.8(30) \\ x^2+6x+9=324 \\ x^2+6x+9-324=0 \\ x^2+6x-315=0 \end{gathered}$

Then, we use the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Where a = 1, b = 6, and c = -315.

$\begin{gathered} x_{1,2}=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-315)}}{2\cdot1} \\ x_{1,2}=\frac{-6\pm\sqrt[]{36+1260}}{2}=\frac{-6\pm\sqrt[]{1296}}{2} \\ x_{1,2}=\frac{-6\pm36}{2} \\ x_1=\frac{-6+36}{2}=\frac{30}{2}=15 \\ x_2=\frac{-6-36}{2}=\frac{-42}{2}=-21 \end{gathered}$ <h2>Hence, the answer is 15 because lengths can't be negative.</h2>

7 0

1 year ago

In spring 2014, faculty from the City University of New York (CUNY) reported on a randomized controlled trial they conducted. Th

Lady_Fox [76]

Answer:

A) explanatory variable is "The extra support given by faculties"

B) the response variable is;

"How successful students who were placed directly into college - level statistics were able to outperform both elementary algebra treatment groups."

C) Confounding Variable is "ability of students to be successful or pass"

D) Experiment

Step-by-step explanation:

A) An explanatory variable is simply a type of independent variable used to explain the changes that occur in another variable. Furthermore, it can be manipulated by the researcher and is also used to predict differences in the response variable.

Now, from the question, we see that the variable "extra support by the faculties" is used to determine how successful community college students who were placed directly into college - level statistics were in outperforming both elementary algebra treatment groups.

Thus, the explanatory variable is "The extra support given by faculties"

B) In answer A above, we saw that the explanatory variable is used to predict the response variable.

We saw that the explanatory variable predicted how successful community college students who were placed directly into college - level statistics were in outperforming both elementary algebra treatment groups.

Thus, the response variable is;

"How successful students who were placed directly into college - level statistics were able to outperform both elementary algebra treatment groups."

C) A confounding variable is one that influences or changes the effect of the dependent & independent variable. It is therefore used to influence the result/outcome of an experiment/research.

From the question, it's clear that the ability of the students to pass or be successful is what influences the effect of the support by the faculties.

Thus, the Confounding Variable is "ability of students to be successful or pass"

D) The question portrays an experiment because it is being used to check if an hypothesis statement is true or false.

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

Suppose a 90% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$38,737, $50,46

JulsSmile [24]

Answer:

a. The point estimate was of $44,600.

b. The sample size was of 16.

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

The margin of error is the difference between the two bounds, divided by 2.

a. What is the point estimate of the mean salary for all college graduates in this town?

Mean of the bounds, so:

(38737+50463)/2 = 44600.

The point estimate was of $44,600.

b. Determine the sample size used for the analysis.

First we need to find the margin of error, so:

$M = \frac{50463-38737}{2} = 5863$

Relating the margin of error with the sample size:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.05 = 0.95$ , so Z = 1.64.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

For this problem, we have that $\sigma = 14300, M = 5863$ . So

$M = z\frac{\sigma}{\sqrt{n}}$

$5863 = 1.645\frac{14300}{\sqrt{n}}$

$5863\sqrt{n} = 1.645*14300$

$\sqrt{n} = \frac{1.645*14300}{5863}$

$(\sqrt{n})^2 = (\frac{1.645*14300}{5863})^2$

$n = 16$

The sample size was of 16.

7 0

3 years ago

Can you help me find x

kifflom [539]

X = 30

hope that helps

180 - 108 = 72

180 - 72 -43 = 65

triangle on the left
Angles = 43,65,72

triangle on the right

180 - 65 = 115

180 - 115 - 35 = 30
so angles of triangle on the right = 30, 35, 115

x = 30 (vertical angles with the other angle of triangle on the right)

7 0

3 years ago