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2 years ago

What 2 numbers add up to be 6 but multiply together to be 4?

Mathematics

2 answers:

slega [8]2 years ago

5 0

4 because 2+4=6 and 4 x 1 = 4

trasher [3.6K]2 years ago

4 0

4,because 2+5 equals to 6 and 4 times one is 4

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The data given below contain the state cigarette taxâ€‹ (in $) for all 20 regions of a country.

g100num [7]

Answer:

(a) $\bar x= 1.583$ -- Population Mean

(b) $s = 1.032$ --- Population standard deviation

Step-by-step explanation:

Given:

Cigarette tax for 20 regions

Solving (a): The population mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

$\sum x = 1.36 + 1.70 + 2.50 + 0.45 + 1.18 + 0.64 + 3.46 + 0.57 + 2.00 + 0.80 + 1.60 +0.98 + 0.36 + 2.24 + 4.35 + 0.62 + 2.70 + 1.78 + 1.53 + 0.84$

$\sum x = 31.66$

$n = 20$

So, we have:

$\bar x= \frac{31.66}{20}$

$\bar x= 1.583$

Solving (b): The population standard deviation

This is calculated as:

$s = \sqrt{\frac{\sum( x - \bar x)^2}{n}$

$\sum (x -\bar x)^2 = (1.36 - 1.583)^2 + (1.70 - 1.583)^2+ (2.50 - 1.583)^2+ (0.45 - 1.583)^2+ (1.18 - 1.583)^2+ (0.64 - 1.583)^2+ (3.46 - 1.583)^2+ (0.57 - 1.583)^2+ (2.00 - 1.583)^2+ (0.80 - 1.583)^2+ (1.60 - 1.583)^2+(0.98 - 1.583)^2+ (0.36 - 1.583)^2+ (2.24 - 1.583)^2+ (4.35 - 1.583)^2+ (0.62 - 1.583)^2+ (2.70 - 1.583)^2+ (1.78 - 1.583)^2+ (1.53 - 1.583)^2+ (0.84- 1.583)^2$

$\sum (x -\bar x)^2 = 21.29222$

So:

$s = \sqrt{\frac{21.2922}{20}$

$s = \sqrt{1.06461}$

$s = 1.032$

Solving (c):

Population mean tells the average amount while the standard deviation represents the spread from the calculated mean

Option (4) is correct

6 0

2 years ago

Please help!!! I don’t understand!

soldier1979 [14.2K]

Answer:

y = 3x

Step-by-step explanation:

Hope this helps!

-Josh

3 0

1 year ago

gave each student a pair of standard dice. The students rolled the dice once and reported their results to the teacher. How many

Salsk061 [2.6K]

42 times is the answer

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2 years ago

Suppose you want to buy one pair of pants and several pairs of socks.

dedylja [7]

Answer:

Step-by-step explanation:

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2 years ago

Report Error Suppose $P(x)$ is a polynomial of smallest possible degree such that: $\bullet$ $P(x)$ has rational coefficients $\

motikmotik

Answer:

We want a polynomial of smallest degree with rational coefficients with zeros in $\sqrt{7}$ , $1 - \sqrt{6}$ and -3. The last root gives us the factor (x+3). Hence, our polynomial is

$P(x) =(x+3)q(x)$

where $q$ is a polynomial with rational coefficients and roots $\sqrt{7}$ and $1 - \sqrt{6}$ . The root $\sqrt{7}$ gives us a factor $x-\sqrt{7}$ , but in order to obtain rational coefficients we must consider the factor $x^2-7$ .

An analogue idea works with $1 - \sqrt{6}$ . For convenience write $x - 1 + \sqrt{6} = ( x - 1) + \sqrt{6}$ . This gives the factor $(x-1)^2-6$ . Hence,

$P(x) = (x+3)(x^2-7)((x-1)^2-6)=x^5+x^4-18x^3-22x^2+77x+105$

Notice that $P(-1)=24$ . So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is

$P(x) =(1/3)x^5+(1/3)x^4-6x^3-(22/3)x^2+(77/3)x+35$

Step-by-step explanation:

We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type $x-\sqrt7$ will introduce in the expression, we need to multiply by its conjugate $x+\sqrt7$ . Hence, we will obtain $x^2-7$ that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.

4 0

2 years ago