All categories

2 years ago

Please Help ASAP (show work if you want its not needed i just need the answer)

Mathematics

1 answer:

cricket20 [7]2 years ago

4 0

$\cos(60) = \frac{12}{x} \\$

$\frac{1}{2} = \frac{12}{x} \\$

$x = 24$

_____________________________________________

$\sin(60) = \frac{y}{x} \\$

$\frac{ \sqrt{3} }{2} = \frac{y}{24} \\$

$y = 12 \sqrt{3}$

You might be interested in

A 30cm ruler sometimes measures inches on the other side.

Alisiya [41]

1inch = 2.5cm

∆ 30cm

5cm =2inch

so 30cm=12inch

4 0

2 years ago

I need help on now!!What is one difference??

Mumz [18]

They're alike because they each have 5 sides. They're different because figure A has no right angles but figure B does.

4 0

3 years ago

Read 2 more answers

Information about the proportion of a sample that agrees with a certain statement is given below. Use StatKey or other technolog

lara31 [8.8K]

Answer:

$SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048$

If we replace the values obtained we got:

$0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443$

The 95% confidence interval would be given by (0.257;0.443)

Step-by-step explanation:

Notation and definitions

$X=35$ number of people that agree.

$n=100$ random sample taken

$\hat p=\frac{35}{100}=0.35$ estimated proportion of people that agree

$p$ true population proportion of peopl that agree

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".

The population proportion have the following distribution

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

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: