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ruslelena [56]
3 years ago
11

PLEASE HELP!!! MATH!!!

Mathematics
2 answers:
labwork [276]3 years ago
3 0
-\frac{1}{2} = y -1\\y=0.5

-3\pi + w = 2\pi\\w= 5\pi

1.2m = 0.6\\m = 0.5

q + 2.7 + -0.9 = q + 1.8

USPshnik [31]3 years ago
3 0

Answer:

Step-by-step explanation:

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Ugo [173]
Yes it is LCM and GCF because you are trying to a factor that they both have
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Find a 90% confidence interval for the mean of all tree ring dates from this archaeological site. (round your answers to the nea
creativ13 [48]
Given a table below which gives the years in A.D. for an archaeological excavation site using the method of tree ring dating:

\begin{tabular}
{|c|c|c|c|c|c|c|c|c|}
1229&1292&1187&1257&1268&1316&1275& 1317&1275
\end{tabular}

Part A:

The sample mean is given by:

\bar{x}= \frac{1229+1292+1187+1257+1268+1316+1275+ 1317+1275}{9} \\  \\  = \frac{11416}{9} =1268

Therefore, the sample mean is 1268.



Part B:

We calculate the sample standard deviation as follows:

s= \frac{1}{9-1} \left(\sqrt{(1229-1268)^2+(1292-1268)^2+(1187-1268)^2+(1257-1268)^2+(1268-1268)^2+(1316-1268)^2+(1275-1268)^2+(1317-1268)^2+(1275-1268)^2}\right)\\ \\ = \frac{1}{8}\left(\sqrt{(-39)^2+24^2+(-81)^2+(-11)^2+0^2+48^2+7^2+49^2+7^2}\right)\\ \\ = \frac{1}{8}\left(\sqrt{1521+576+6561+121+2304+49+2401+49}\right)\\ \\ = \frac{1}{8}\left(\sqrt{13,582}\right)\\ \\ = \frac{1}{8}(116.54)=14.57\approx15



Part C:

The 90% confidence interval is given by:

90\% \ C. \ I.=1268\pm1.65\left(\frac{15}{\sqrt{9}}\right) \\  \\ =1268\pm\left(\frac{15}{3}\right)=1268\pm5 \\  \\ =(1268-5, \ 1268+5)=(1263, \ 1273)
8 0
3 years ago
Solve the trigonometric equation such that 0 ≤ x ≤ 2????. Round to three decimal places.
raketka [301]

Answer: x= 0.463

Step-by-step explanation:

Given the equation √5 cos(x) − 2 = 0

To make 'x' the subject of the formula

√5cosx = 2

Cosx = 2/√5

Cosx = 0.89

x = arccos 0.89

x = 0.463

5 0
3 years ago
If there are x teams in a sports​ league, and all the teams play each other​ twice, a total of​ n(x) games are​ played, where ​n
EleoNora [17]
1.
Consider a group of n objects. Assume we want to form groups of r, from these n objects.

There are in total C(n, r)= \frac{n!}{r!(n-r)!} many ways of doing so.

where, r! is "r factorial", calculated as 1*2*3*...*(r-1)*r

2.

C(x, 2) is the total number of pairs out of x objects, that we can form.

so let the x objects represent the x teams, and 2, represent a group of 2, which means a game.

The total number of games is:

n(x)=2*C(x, 2) = 2*  \frac{x!}{2!(x-2)!}= \frac{x!}{(x-2)!}= \frac{x(x-1)(x-2)!}{(x-2)!}=x(x-1)= x^{2} -x

Remark, we multiplied by 2, since there will be 2 matches for each pair of teams.


Answer: x^{2} -x

6 0
3 years ago
For any nonnegative real number.
frutty [35]

Answer:

A

Step-by-step explanation:

We know that \sqrt{b}=b^{1/2}.

We also know that (a^b)^c=a^{bc}.

We can combine these facts to get (\sqrt{b})^2=(b^{1/2})^2=b^{1/2*2}=b^{1}=b.

So, the answer is \boxed{A} and we're done!

<em>Note: the non-negative part or the equation comes from the fact that you assume the square root of a negative number doesn't exist. You will change this assumption later in the year, or in further years.</em>

5 0
3 years ago
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