Find the measure of angle a.

zhannawk [14.2K]

2/3 (20) (9)/7 You first do the partistis then you simmplfy hope it worked!

8 0

3 years ago

A movie Theater charges $6.50 for morning tickets and $8.75 for night tickets the theater sold 378 tickets for a total amount of

Xelga [282]

Answer:

The number of sold morning tickets are 168.

Step-by-step explanation:

Given,

Total number of tickets = 378

Total earned amount = $2929.50

Charge of each morning ticket = $6.50

Charge of each night ticket = $8.75

Solution,

Let the number of morning tickets be 'x'.

And also let the number of night tickets be 'y'.

So, the total number of tickets is the sum of number of morning tickets and number of evening tickets.

So, framing in equation form, we get;

$x+y=378\ \ \ \ equation\ 1$

Again, Total earned amount is the sum of number of morning tickets multiplied with charge of each ticket and number of evening tickets multiplied with charge of each ticket.

So, framing in equation form, we get;

$6.50x+8.75y=2929.50\ \ \ equation\ 2$

On multiplying equation 2 by 100 on both side, we get;

$100(6.50x+8.75y)=2929.50\times100\\\\650x+875y=292950\ \ \ \ equation\ 3$

Now multiplying equation 1 by '650', we get;

$650(x+y)=378\times650\\\\650x+650y=245700\ \ \ \ equation\ 4$

Now subtracting equation 4 from equation 3, we get;

$(650x+875y)-(650x+650y)=292950-245700\\\\650x+875y-650x-650y=47250\\\\225y=47250\\\\y=\frac{47250}{225}=210$

Now substituting the value of'y' in equation 1, we get;

$x+y=378\\\\x+210=378\\\\x=378-210=168$

Hence The number of sold morning tickets are 168.

Whereas The number of sold night tickets are 210.

5 0

3 years ago

-El papá de Susana es carpintero y en su trabajo utiliza clavos y tornillos de distintas medidas. Ahora, le advirtieron que dent

irinina [24]

Answer:

la tabla no esta, asi que busque la pregunta por interenet y encontre lo siguiente:

medida de los clavos equivalencia en

en pulgadas centimetros

1 pulgada 2.54 cms

¹/₂ pulgada 2.54 x 0.5 = 1.27 cms

³/₄ pulgada 2.54 x 0.75 = 1.905 cms

¹/₈ pulgada 2.54 x 0.125 = 0.3175 cms

3¹/₂ pulgadas 2.54 x 3.5 = 8.89 cms

7 0

3 years ago

The time a randomly selected individual waits for an elevator in an office building has a uniform distribution with a mean of 0.

Amiraneli [1.4K]

Answer:

The mean of the sampling distribution of means for SRS of size 50 is $\mu = 0.5$ and the standard deviation is $s = 0.0409$

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size, of at least 30, can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 0.5, \sigma = 0.289$

What are the mean and standard deviation of the sampling distribution of means for SRS of size 50?

By the Central Limit Theorem

$\mu = 0.5, s = \frac{0.289}{\sqrt{50}} = 0.0409$

The mean of the sampling distribution of means for SRS of size 50 is $\mu = 0.5$ and the standard deviation is $s = 0.0409$

Does it matter that the underlying population distribution is not normal?

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

What is the probability a sample of 50 people will wait longer than 45 seconds for an elevator?

We have to use 45 seconds as minutes, since the mean and the standard deviation are in minutes.

Each minute has 60 seconds.

So 45 seconds is 45/60 = 0.75 min.

This probability is 1 subtracted by the pvalue of Z when X = 0.75. So

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

By the Central Limit Theorem

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