4 percent of a number is what fraction of that number?? I REALLY NEED HELP ASAP

The property allows 3i + 2i to be written as (3 +2)I.

Aleks04 [339]

Answer:

distributive

Step-by-step explanation:

5 0

3 years ago

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The binomial x-2 is a factor of the polynomial below. f(x)=x3+x2+nx+10 What is the value of n?

Nutka1998 [239]

<h3>Answer: n = -11</h3>

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Explanation:

Since x-2 is a factor of f(x), this means f(2) = 0.

More generally, if x-k is a factor of p(x), then p(k) = 0. This is a special case of the remainder theorem.

So if we plugged x = 2 into f(x), we'd get

f(x) = x^3+x^2+nx+10

f(2) = 2^3+2^2+n(2)+10

f(2) = 8+4+2n+10

f(2) = 2n+22

Set this equal to 0 and solve for n

2n+22 = 0

2n = -22

n = -22/2

n = -11 is the answer

Therefore, x-2 is a factor of f(x) = x^3+x^2-11x+10

Plug x = 2 into that updated f(x) function to find....

f(x) = x^3+x^2-11x+10

f(2) = 2^3+2^2-11(2)+10

f(2) = 8+4-22+10

f(2) = 0

Which confirms our answer.

3 0

4 years ago

7 divided by 1/2 <img src="https://tex.z-dn.net/?f=7%20%5Cdiv%7C%201" id="TexFormula1" title="7 \div| 1" alt="7 \div| 1" alig

Elan Coil [88]

The answer should be 14

7 0

2 years ago

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The probability that a person in the united states has type b+ blood is 13%. three unrelated people in the united states are s

zubka84 [21]

$\frac{13}{100} \times \frac{13}{100} \times \frac{13}{100} \\ = \frac{2197}{1000000}$
Answer is 0.2197%

Hope this helps. - M

6 0

3 years ago

The table showing the stock price changes for a sample of 12 companies on a day is contained in the Excel file below.

AfilCa [17]

Answer:

(a) The sample variance for the daily price change is 0.2501.

(b) The sample standard deviation for the daily price change is 0.5001.

(c) The 95% confidence interval estimates of the population variance is (0.1255, 0.7210).

Step-by-step explanation:

Let the random variable X denote the stock price changes for a sample of 12 companies on a day.

The data provided is:

X = {0.82 , 1.44 , -0.07 , 0.41 , 0.21 , 1.33 , 0.97 , 0.30 , 0.14 , 0.12 , 0.42 , 0.15}

(a)

The formula to compute the sample variance for the daily price change is:

$s^{2}=\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}$

The sample mean is computed using the formula:

$\bar X=\frac{1}{n}\sum\limits^{12}_{i=1}{X_{i}}$

Consider the Excel output attached below.

In Excel the formula to compute the sample mean and sample variance are:

$\bar X$ =AVERAGE(A2:A13)

$s^{2}$ =VAR.S(A2:A13)

Thus, the sample variance for the daily price change is 0.2501.

(b)

The formula to compute the sample standard deviation for the daily price change is:

$s=\sqrt{\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}}$

Consider the Excel output attached below.

In Excel the formula to compute the sample standard deviation is:

$s$ =STDEV.S(A2:A13)

Thus, the sample standard deviation for the daily price change is 0.5001.

(c)

The (1 - α)% confidence interval for population variance is:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

Compute the critical value of Chi-square for α = 0.05 and (n - 1) = (12 - 1) = 11 degrees of freedom as follows:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2,11}=21.920$

$\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{(1-0.05/2),11}=\chi^{2}_{0.975,11}=3.816$

*Use a Chi-square table.

Compute the 95% confidence interval estimates of the population variance as follows:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

$=[\frac{(12-1)\times 0.2501}{21.920 } \leq \sigma^{2}\leq \frac{(12-1)\times 0.2501}{3.816} ]$

$=[0.125506\leq \sigma^{2}\leq 0.720938]\\\approx [0.1255, 0.7210]$

Thus, the 95% confidence interval estimates of the population variance is (0.1255, 0.7210).

7 0

4 years ago