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2 years ago

The question is in the attachment. Please answer

Mathematics

2 answers:

maria [59]2 years ago

4 0

The guy on top is correct

Temka [501]2 years ago

4 0

Answer:

74°

Step-by-step explanation:

LMN is a central angle so LMN = LN

LN = 74°

LMN = 74°

You might be interested in

An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0

zmey [24]

Answer:

96.33% probability that everyone who appears for the flight will get a seat

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 105, p = 0.9$

$\mu = E(X) = np = 105*0.9 = 94.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{105*0.9*0.1} = 3.07$

What is the probability that everyone who appears for the flight will get a seat

100 or less people appearing to the flight, which is the pvalue of Z when X = 100. So

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

$Z = \frac{100 - 94.5}{3.07}$

$Z = 1.79$

$Z = 1.79$ has a pvalue of 0.9633

96.33% probability that everyone who appears for the flight will get a seat

7 0

2 years ago

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$

$Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)$

$(Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}$

$Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}$

$Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}$

$n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2$

$n = (16*(\frac{1.2}{(270-264)}))^2$

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

6 0

2 years ago

Write 104/-120 as a rational number in its standard form

AVprozaik [17]

104/-120

=-13/15

Devishri1977 gave a better example of how she got to the answer

5 0

2 years ago

Read 2 more answers

Which of the following represents the equation of the line passing through (-4,5)

Sergio [31]

Answer:

Answer: y=2x+13.

Step-by-step explanation:

Your input: find the equation of the line perpendicular to the line y=5/2−x/2 passing through the point (−4,5).

The equation of the line in the slope-intercept form is y=5/2−x/2.

The slope of the perpendicular line is negative inverse: m=2.

So, the equation of the perpendicular line is y=2x+a.

To find a, we use the fact that the line should pass through the given point: 5=(2)⋅(−4)+a.

Thus, a=13.

Therefore, the equation of the line is y=2x+13.

Answer: y=2x+13.

5 0

3 years ago

two trains leave the station at the same time one heading west and the other East the westbound train travels at 55 miles per ho

Diano4ka-milaya [45]

Recall your d = rt, distance = rate * time

so... one train goes west and the goes the opposite way... alrite... so... notice, by the time 338 miles have been covered by both, it will have been "t" hours, and whatever "t" is, is the same amount of time the westbound train has been running as well as the eastbound train has been running.

now, let's say, since by "t" hours they've covered 338 altogether, so, if the westbound train has covered say "d" miles, then the eastbound train would have covered the slack from 338 and d, that is, "338 - d".

$\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
Westbound&d&55&t\\
Eastbound&338-d&75&t
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=55t\\
338-d=75t\\
----------\\
338-\boxed{55t}=75t
\end{cases}
\\\\\\
338=75t+55t\implies 338=130t\implies \cfrac{338}{130}=t\implies \stackrel{hours}{2.6}=t$

so, 2hours and 36 minutes.

8 0

3 years ago