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nekit [7.7K]
3 years ago
6

Which survey is most likely affected by bias

Mathematics
1 answer:
Semenov [28]3 years ago
8 0
I would guess C, but I am not sure
The reason for this is because some students might strongly believe you should be allowed to have cell phones, but others are strongly against it
Sorry if I am wrong
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Identify the slope and y-intercept: y = -x + 13 y help mee please​
Marina86 [1]

Answer:

slope : -1, y-intercept : 13

Step-by-step explanation:

-x = -1 times x

+13 = 13 = y-intercept

7 0
3 years ago
Read 2 more answers
The perimeter of a rectangular field is 340 yards. The length is 90 yd longer than the width. Find the dimensions.
Alexus [3.1K]

Answer:

length= 100 width=70

Step-by-step explanation:

The perimeter of a rectangular field is 340 yds.  

The length is 30 yds longer than the width.  

2(l + w) = 340

l + w = 170

w + 30 + w = 170

2w = 140

w = 70

Hence the dimensions:

Length: 100 yds.

Width: 70 yds.

8 0
3 years ago
Read 2 more answers
5
frutty [35]

Answer:

she started with 50 marbles

Step-by-step explanation:

The total is 100 than you just subtract 20 and 30 becasue that is what is added later on. After you subtract 20 and 30 from 100 you will get 50. Therefore Sara had started out with 50 marbles.

Hope that helps :)

6 0
3 years ago
Read 2 more answers
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
FromTheMoon [43]

Answer:

The Taylor series is \ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

The radius of convergence is R=3.

Step-by-step explanation:

<em>The Taylor expansion.</em>

Recall that as we want the Taylor series centered at a=3 its expression is given in powers of (x-3). With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of \ln(1+x).

Then,

\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).

Now, in order to make a more compact notation write \frac{x-3}{3}=y. Thus, the above expression becomes

\ln(x) = \ln 3 + \ln(1+y).

Notice that, if x is very close from 3, then y is very close from 0. Then, we can use the Taylor expansion of the logarithm. Hence,  

\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.

Now, substitute \frac{x-3}{3}=y in the previous equality. Thus,

\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

<em>Radius of convergence.</em>

We find the radius of convergence with the Cauchy-Hadamard formula:

R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},

Where a_n stands for the coefficients of the Taylor series and R for the radius of convergence.

In this case the coefficients of the Taylor series are

a_n = \frac{(-1)^{n+1}}{ n3^n}

and in consequence |a_n| = \frac{1}{3^nn}. Then,

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}

Applying the properties of roots

\sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.

Hence,

R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}

Recall that

\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.

So, as R^{-1}=\frac{1}{3} we get that R=3.

8 0
4 years ago
The triangles below are not drawn to scale or to proper proportions.
Lesechka [4]

Answer/Step-by-step explanation:

Given:

m<A = 115,

m<B = 78,

m<C = 35,

✔️m<1 = 180 - m<A (angles on a straight line/supplementary angles)

m<1 = 180 - 115 (substitution)

m<1 = 65°

✔️m<2 + m<1 + m<B = 180° (sum of interior angles of a triangle)

m<2 + 65 + 78 = 180

m<2 + 143 = 180

m<2 = 180 - 143 (substitution property of equality)

m<2 = 37°

✔️m<3 = 180° - m<B (supplementary angles)

m<3 = 180 - 78

m<3 = 102°

4 0
3 years ago
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