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2 years ago

Find y.Help me please:))

Mathematics

2 answers:

svlad2 [7]2 years ago

8 0

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \: y=47 \degree$

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$\large \tt Solution \: :$

$\qquad \tt \rightarrow \:(2y - 3) + 50 = 3y$

$\qquad \tt \rightarrow \:2y - 3 + 50 - 3y = 0$

$\qquad \tt \rightarrow \:2y - 3y + 47 = 0$

$\qquad \tt \rightarrow \: - y = - 47$

$\qquad \tt \rightarrow \:y = 47 \degree$

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

Lesechka [4]2 years ago

5 0

Answer:

y = 47

Step-by-step explanation:

All of the angles in a triangle must add up to 180. The angle marked 3y can help us find the missing angle of the triangle: 180-3y
Then you can make an equation to help solve for y:
50 + (2y-3) + (180-3y) = 180
Combine like terms:
227-y=180
y = 47

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A student was asked to find a 90% confidence interval for the proportion of students who take notes using data from a random sam

Scorpion4ik [409]

Answer:

After use the formula we got the following result for the 90% confidence interval (0.13 <p<0.34)

And the conclusion for this case would be:

e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.

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".

Description in words of the parameter p

$p$ represent the real population proportion of students who take notes

$\hat p$ represent the estimated proportion of students who take notes

n is the sample size required

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

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

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

Confidence interval

The confidence interval for a proportion is given by this formula

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

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.