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3 years ago

Domain: Ranges: Zeros: Discontinuities: Asymptotes:

Mathematics

1 answer:

trapecia [35]3 years ago

4 0

Domain: The set of possible values of x

Range: The set of possible values of y

Zeros: The x-intercepts (what value of x is on the x-axis)

Discontinuities: Where the graph stops (and restarts somewhere else)

Asymptotes: The graph cannot cross this line (vertical, horizontal, oblique)

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In an election to the U.S. Senate, one candidate has popular support of 53% and the other has support of 47%. On election eve, a

Irina18 [472]

Answer:

a) 94.63% probability that the poll will predict the correct winner.

b) 69.15% probability of correct prediction. The lesser sample size leads to more variation, which means that the probability of correct prediction is smaller than in (a).

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

(a) Compute a normal approximation with continuity correction of the probability that the poll will predict the correct winner.

The correct winner is the one with 53% of the votes.

We have that $p = 0.53, n = 750$ .

$E(X) = np = 750*0.53 = 397.5$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{750*0.53*0.47} = 13.67$

The pool will correctly predict the winner if the candidate that has a support of 53% gets more than half the votes.

So we have to find P(X > 750/2) = P(X > 375)[/tex]

Using continuity correction, this is P(X > 375 + 0.5) = P(X > 375.5), which is 1 subtracted by the pvalue of Z when X = 375.5. So

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

$Z = \frac{375.5 - 397.5}{13.67}$

$Z = -1.61$

$Z = -1.61$ has a pvalue of 0.0537

1 - 0.0537 = 0.9463

94.63% probability that the poll will predict the correct winner.

(b) What happens to the probability of correct prediction if they had conducted the poll over only 100 voters?

Lesser sample size leads to more variation, which means that the probability of correct prediction will be smaller.

Now $n = 100$

Then

$E(X) = np = 100*0.53 = 53$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.53*0.47} = 4.99$

Also P(X > 50), which using continuity correction is P(X > 50.5).

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

$Z = \frac{50.5 - 53}{4.99}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3085

1 - 0.3085 = 0.6915.

69.15% probability of correct prediction. The lesser sample size leads to more variation, which means that the probability of correct prediction is smaller than in (a).

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3 years ago

Mario invested $5100 $ ⁢ 5100 at 13% 13 % to be compounded daily. What will be the value of Mario's investment in 2 2 years? Rou

Gennadij [26K]

Answer:

lets see

Step-by-step explanation:

5 0

3 years ago

Find a z-score that has a distribution less than 30.5%

Ludmilka [50]

To the right: 0.51
to the left: -0.51

8 0

3 years ago

Read 2 more answers

HELP ME I WILL GIVE BRAINLIEST

Nostrana [21]

Answer:

8760 hours

Step-by-step explanation:

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3 years ago

Reinhardt Furniture Company has 40,000 shares of cumulative preferred 2% stock, $150 par and 100,000 shares of $5 par common sto

Ann [662]

Answer:

for year 1

common stock = $1.75 per share

preferred stock = Zero

for year 2

common stock = $4.25 per share

preferred stock = $0.3 per share

for year 3

common stock = $3 per share

preferred stock = $2 per share

Step-by-step explanation:

step 1

preferred stock value = (40000 shares * $150) = $6000000

common stock value = (100000 shares * $5) = $500000

step 2

For year 1:

Dividend on preferred stock;

$\frac{6000000 * 2}{100}$ = $120000

But total dividend in the question was $70000 therefore total amount of dividend on cumulative preferred stock is $70000.

hence, dividend per share

$= \frac{70000}{40000 shares}$ = $1.75 per share

Dividend on common stock;

70,000 - 70,000 = Zero

as total dividend distributed in year 1 is insufficient for cumulative preferred stock therefore no dividend will be paid on common stock.

For year 2:

Dividend on cumulative preferred stock;

$\frac{6000000 * 2}{100}$ = $120000

extra dividend of year 1 ($120000 - $70000) = $50000

Thus total dividend on cumulative preferred stock

($120000 + $50000) = $170000

So dividend per share

$\frac{170000}{40000\ shares}$ = $4.25 per share

Dividend on common stock;

($200000 – $170000) = $30000

dividend per share

$\frac{30000}{100000\ shares}$ = $0.3 per share

For year 3:

Dividend on cumulative preferred stock;

$\frac{6000000 * 2}{100}$ = $120000

total dividend on cumulative preferred stock $120000

dividend per share

$\frac{120000}{40000 shares}$ = $3 per share

No dividend was extra in the year 2 therefore only available dividend of this year will be paid.

Dividend on common stock;

($320000 – $120000) = $200000

dividend per share

$\frac{200000}{100000\ shares}$ = $2 per share

3 0

3 years ago