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worty [1.4K]
3 years ago
11

35 - 8 break apart ones to subtract

Mathematics
1 answer:
noname [10]3 years ago
4 0
The. Answer is 27 I don’t get what brake apart is
You might be interested in
A poll in 2017 reported that 705 out of 1023 adults in a certain country believe that marijuana should be legalized. When this p
just olya [345]

Answer:

1. d. (0.652, 0.726)

2. b. (0.661, 0.718)

a. The margin of error of a 90​% confidence interval will be less than the margin of error for the 95​% and 99​% confidence intervals because intervals get wider with increasing confidence level.

Step-by-step explanation:

Data given and notation  

n=1023 represent the random sample taken in 2017    

X=705 represent the people who thinks that believe that marijuana should be legalized.

\hat p =\frac{705}{1023}=0.689 estimated proportion of people who thinks that believe that marijuana should be legalized.

z would represent the statistic in order to find the confidence interval    

p= population proportion of people who thinks that believe that marijuana should be legalized.

Part 1

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.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.689 -2.58 sqrt((0.689(1-0.689))/(1023))=0.652

0.689 + 2.58sqrt((0.56(1-0.689))/{1023))=0.726

And the 99% confidence interval would be given (0.652;0.726).

We are 99% confident that about 65.2% to 72.6% of people  believe that marijuana should be legalized

d. (0.652, 0.726)

Part 2

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 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.689 - 1.96((0.689(1-0.689))/(1023))=0.661

0.689 + 1.96 ((frac{0.56(1-0.689))/(1023))=0.718

And the 95% confidence interval would be given (0.661;0.718).

We are 95% confident that about 66.1% to 71.8% of people  believe that marijuana should be legalized

b. (0.661, 0.718)

Part 3

Would be lower since the quantile z_{\alpha/2} for a lower confidence is lower than a quantile for a higher confidence level.

The margin of error of a 90​% confidence interval will be less than the margin of error for the 95​% and 99​% confidence intervals

If we calculate the 90% interval we got:

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution. The quantile for this case would be 1.64.

And replacing into the confidence interval formula we got:

0.689 - 1.64 ((0.689(1-0.689))/{1023))=0.665

0.689 + 1.64 ((0.56(1-0.689))/(1023))=0.713

And the 90% confidence interval would be given (0.665;0.713).

We are 90% confident that about 66.5% to 71.3% of people  believe that marijuana should be legalized.

The intervals get wider with increasing confidence level.

So the correct answer is:

a. The margin of error of a 90​% confidence interval will be less than the margin of error for the 95​% and 99​% confidence intervals because intervals get wider with increasing confidence level.

6 0
3 years ago
C<br> On Wednesday they made $5.50 How much does Zara get?<br><br> Please help?
Natalija [7]

Answer

Zara gets $5.50

Step-by-step explanation:

7 0
2 years ago
Help please thank you
LuckyWell [14K]

Answer:

hmm i think its

1/2(x + y)

Step-by-step explanation:

because well... this would be eaiser for me if it said "of the" because "of the" is mostly used for multiply, but since it just said "the..." i can only guess. I think this may be the answer tho?

7 0
1 year ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B
riadik2000 [5.3K]

Let

S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n

where we assume |r| < 1. Multiplying on both sides by r gives

r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}

and subtracting this from S_n gives

(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}

As n → ∞, the exponential term will converge to 0, and the partial sums S_n will converge to

\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}

Now, we're given

a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a

a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}

We must have |r| < 1 since both sums converge, so

\dfrac{15}a = \dfrac1{1-r}

\dfrac{150}{a^2} = \dfrac1{1-r^2}

Solving for r by substitution, we have

\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)

\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}

Recalling the difference of squares identity, we have

\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to

\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15

It follows that

\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is

ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0
2 years ago
Hi, please help me :)
bearhunter [10]

Answer:

D. Linear

The answer is negative linear, because when it incresing from 0 to 1 to 2 to 3 (year) it always deacreasing in the value($).

4 0
3 years ago
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