Ask question

Ask question

All categories

2 years ago

9

Pythagorean theorem round to the nearest hundredth

2 answers:

Katyanochek1 [597]2 years ago

5 0

Answer:

x = 14.97

Step-by-step explanation:

a= 10 c=18

1. plug variables in for pythagorean theorem.

$a^2+b^2=c^2$ → $10^2+b^2=18^2$

2.solve for b from pythagorean theorem.

$10^2+b^2=18^2$ → $100+b^2=324$

-100 -100

$b^2=224$ → $b=\sqrt{224}$

b = 14.9666295.....

3. Round variable b to the nearest hundreth assuming they expect that and change the variable to x since we were actually looking for x.

b=14.9666....→ b(estimated)= 14.97

x(estimated)=14.97

Ber [7]2 years ago

5 0

The answer your going To end up with is x=14.97

You might be interested in

In a binomial experiment with 45 trials, the probability of more than 25 success can be approximated by What is the probability

Pani-rosa [81]

Answer:

0.6 is the probability of success of a single trial of the experiment

Complete Problem Statement:

In a binomial experiment with 45 trials, the probability of more than 25 successes can be approximated by $P(Z>\frac{(25-27)}{3.29})$

What is the probability of success of a single trial of this experiment?

Options:

0.07
0.56
0.79
0.6

Step-by-step explanation:

So to solve this, we need to use the binomial distribution. When using an approximation of a binomially distributed variable through normal distribution , we get:

$\mu =\frac{np}{\sigma}$ = $\sqrt{np(1-p)}$

now,

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

so,

by comparing with $P(Z>\frac{(25-27)}{3.29})$ , we get:

μ=np=27

$\sigma=\sqrt{np(1-p)}$ =3.29

put np=27

we get:

$\sigma=\sqrt{27(1-p)}$ =3.29

take square on both sides:

10.8241=27-27p

27p=27-10.8241

p=0.6

Which is the probability of success of a single trial of the experiment

5 0

3 years ago

A dozen plain hats cost 3.60,what is the unit rate of a hat?

deff fn [24]

I think the answer should be 12.

7 0

3 years ago

Read 2 more answers

Which of the following statements is not true? A. If the probability of an event occurring is 1.5, then it is certain that eve

yawa3891 [41]

Answer:

Therefore, we conclude that the statement in (A) is incorrect.

Step-by-step explanation:

We have the following sentences:

A) If the probability of an event occurring is 1.5, then it is certain that event will occur.

B) If the probability of an event occurring is 0, then it is impossible for that event to occur.

We know that the range of probability of an event occurring is in the segment [0, 1]. In statement under (A), we have the probability that is equal to 1.5.

Therefore, we conclude that the statement in (A) is incorrect.

8 0

3 years ago

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

<u />

<u>(a) The books in any order</u>

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

<u>(b) The mathematics book, not together</u>

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

<u>(c) The novels must be together and the chemistry books, together</u>

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

<u>(d) The mathematics must be together and the chemistry books, not together</u>

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$

Read more about permutations at:

brainly.com/question/1216161

8 0

2 years ago

In which direction must the graph of f(x) = |x| be shifted to produce the graph of g(x) = |x + 4|?

Tamiku [17]

Answer:

Left on the x-axis

Step-by-step explanation:

The opposite of the coefficient's amount of units is how much the graph will be moved.

Table comparisons

g(x)=|x+4|

x 1 -2 <u>-4</u>

y 5 2 <u>0</u>

Underlined is the x-int of the equation.

f(x)=|x|

x 1 <u>0</u> -1 3

y 1 <u>0</u> 1 3

5 0

3 years ago

Read 2 more answers