All categories

3 years ago

What is the value of y?

Mathematics

1 answer:

ddd [48]3 years ago

6 0

Answer:

Step-by-step explanation:

180- half of 42

138= x+ measure of 9

so means

10+6=16means other side is also 16

7=y+x

an arc is half the inner measure

so really 84-42

hope this helps

You might be interested in

the sum of the measures of the interior angles of the parallelogram is 360° write and solve an equation to find the value of the

AlekseyPX

Dividing the parallelogram in half and applying trigonometry is the easiest solution for me.

4 0

2 years ago

The mean life of a television set is 119 months with a standard deviation of 14 months. If a sample of 74 televisions is randoml

irina [24]

Answer:

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

Step-by-step explanation:

To solve this question, 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 $\sigma$ , the sample means with size n 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:

$\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63$

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 months

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

$Z = \frac{120.1 - 119}{1.63}$

$Z = 0.68$

$Z = 0.68$ has a pvalue of 0.7517

X = 117.9

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

$Z = \frac{117.9 - 119}{1.63}$

$Z = -0.68$

$Z = -0.68$ has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

8 0

2 years ago

Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 produ

Allisa [31]

Answer:

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

Step-by-step explanation:

There is a random binomial variable $X$ that represents the number of units come off the line within product specifications in a review of $n$ Bernoulli-type trials with probability of success $0.91$ . Therefore, the model is ${15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}$ . So: