All categories

3 years ago

Which survey question is biased?

Mathematics

2 answers:

VladimirAG [237]3 years ago

7 0

Answer:

all the questions biased; but if you have choose one it would A.

Step-by-step explanation:

Leni [432]3 years ago

6 0

Answer:

Step-by-step explanation:

It was trick question because they all are opinionated, but the first one should be an easy yes or no question. However, if you do not like the president then you would say no because of bias. The rest are supposed to be "biased" because they are asking what you prefer.

You might be interested in

The sales tax on a used car is $168168, and the sales tax rate is 55%. Find the purchase price (the price before taxes are

Naily [24]

Answer:

$92492.40

Step-by-step explanation:

168168.00 - 55% (75675.60)= $92492.40

7 0

4 years ago

Estimate 3/8 + 3/4 + 7/8 =

goldfiish [28.3K]

Answer:

Step-by-step explanation:

3/8+3/4+7/8

=3/8+6/8+7/8

=16/8

8 0

2 years ago

1. How many squares will it take<br> to cover the figure below?

Ne4ueva [31]

Answer:

Unable to tell you

Step-by-step explanation:

Due to the fact we can't see the shape below, we are unable to help you. make sure to post a picture next time!

8 0

4 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago

Write an algebraic expression to represents the verbal expression: the difference between the product of 4 and a number and the

Naily [24]

Answer:

$4n-n^2$

Step-by-step explanation:

I use 'n' as the unknown value, but you can use any letter you want as the variable.

The product of four and a number would be represented by '4 * n'. The ideal way to write this would be '4n', even though '4 * n' is also correct in value.

The square of the number would be n². Squaring a number is multiplying a number by itself.

The phrase 'difference between...' means subtraction.

Putting it together:

"The difference between the product of 4 and a number and the square of the number" would be:

$4n-n^2$

Hope this helps.

4 0

3 years ago