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3 years ago

Find the value of x. Round your answer to the nearest tenth.

Mathematics

1 answer:

xxMikexx [17]3 years ago

3 0

So for this, we will be using the pythagorean theorem, which is $leg^2+leg^2=hypotenuse^2$ . 9 is being used instead of 18 because we are using only one half of this triangle.

Firstly, apply the numbers to their place $9^2+20^2=x^2$

Solve the exponents to get $81+400=x^2$

Add up the left half of the equation to get $481=x^2$

And lastly, square root both sides of the equation, and your answer should be $21.9=x$

You might be interested in

Sarah walks to school, 1.5 miles in 30 minutes. what is the snswer to this

nevsk [136]

Answer: 3 mi/h

Step-by-step explanation:

Here we have to find the speed $V$ at which Sarah walks to school, knowing the distance $d$ she has walked and the time $t$ she expends.

So, in this case we can use the following equation:

$V=\frac{d}{t}$

Where:

$d=1.5 mi$

$t=30 min \frac{1 h}{60 min}=0.5 h$

Solving:

$V=\frac{1.5 mi}{0.5 h}$

$V=3 mi/h$ This is Sarah's speed

5 0

3 years ago

There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

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".

The population proportion have the following distribution

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

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

The confidence interval would be given by this formula

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

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.