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3 years ago

13

which explains the correct way to move the decimal point to find the quotient in the problem 12.46 divided by 100? My choices ar

e; (A) 2 places to the left (B) 2 places to the right (C) 3 places to the left (D) 3 places to the right?

1 answer:

jarptica [38.1K]3 years ago

6 0

Well, since it has 2 zeros, you move the point 2 places to the right.

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Napoleon went on a bus trip traveling at a speed of 55 miles per hour for 4 hours which equation will help you find the distance

azamat

Answer:

220

Step-by-step explanation:

Easy!

Multiply 55 by 4!

2

55

x 4

--------

220

five times 4 is 20, so put 0 then regroup the 2. Then, 5 times 4 again, but add two!

3 0

3 years ago

Aiden has $15.00 on his copy card. Each time he uses the card to make a photocopy, $0.06 is deducted from his card. Aiden wants

melamori03 [73]

15 - 0.06X=5
would be
X=166 2/3

4 0

4 years ago

Read 2 more answers

Find the measure of the indicated angle to the nearest degree

Y_Kistochka [10]

Answer:

32°

Step-by-step explanation:

use this formula :

Now you want to get the angle be x, so the opposite side is 28 and hypotenuse is 53. Putting in the formula we get:

Sin(x) = 28/53

Sin(x)= 0.5283018868

So to find x just do Sin^-1

x = Sin^-1 (0.5283018868)

you get 31.8907918, ok so we just want the nearest degree and if we do that we get 32˚

if you put them into your calculator you will get 32˚ as your answer

5 0

3 years ago

The shorter leg of a right triangle is 21 feet less than the other leg. Find the length of the two legs if the hypotenuse is 39

mina [271]

Answer:

leg 1= 36 feet

leg2 = 15 feet

Step-by-step explanation:

Hi, we have to apply the Pythagorean Theorem:

c^2 = a^2 + b^2

Where c is the hypotenuse of the triangle (in this case the distance between Doreen’s house and the tower) and a and b are the other legs.

leg1 = x

leg2 = x-21 (21 feet less than the other leg)

Replacing with the values given:

39^2 = x^2 + (x-21)^2

1,521 = x^2 + x^2 -42x +441

0 = 2x^2 -42x +441-1,521

0= 2x^2 -42x -1,080

For: ax2+ bx + c

x =[ -b ± √b²-4ac] /2a (quadratic formula)

Replacing with the values given:

x=-(-42)± √(-42)²-4(2)-1080] /2(2)

x= 42± √10,404] /4

x = 42± 102 /4

Positive:

x = 42+102 /4 = 36

leg 1= 36

leg2 = 36-21 =15

Feel free to ask for more if needed or if you did not understand something.

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 <em>X</em> denote the stock price changes for a sample of 12 companies on a day.

The data provided is:

<em>X</em> = {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 - <em>α</em>)% 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 <em>α</em> = 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