All categories

3 years ago

What is the range for the following numbers? 5, 1, 12, 3, 4, 2, 10

Mathematics

2 answers:

Soloha48 [4]3 years ago

7 0

The range of a data set is the largest number - the smallest number.

In this problem,

The largest number = 12

The smallest number = 1

12 - 1 = 11

The range = 11

kogti [31]3 years ago

4 0

Answer:

Step-by-step explanation:

10 - 5 = 5 the range is the difference from the highest value to the lowest value

You might be interested in

A survey was conducted to measure the height of men. In the survey, respondents were grouped by age. In the 20-29 group the grou

kotykmax [81]

Answer:

(a) The probability that his height is less than 66 inches is 0.2743.

(b) The probability that the height is between 66 and 71 inches is 0.4679.

Step-by-step explanation:

The data given in the question is:

Mean (μ) = 68.4

Standard Deviation (σ) = 4

Let X denote the height of men. We will use the normal distribution z-score formula to calculate the z-score and then look up the probability in the normal probability distribution table. The z-score formula is:

z = (X - μ)/σ

(a) For P(X<66), first calculate the z value.

z = (66-68.4)/4

z = -0.6 (Look up this value in the standard normal distribution table)

P(z<-0.6) = 0.2743

The probability that his height is less than 66 inches is 0.2743.

(b) P(66<X<71) = P(X<71) - P(X<66)

We need to find P(X<71) so, calculating the z-value:

z = (71-68.4)/4

z = 0.65

P(z<0.65) = 0.7422

P(66<X<71) = 0.7422 - 0.2743

P(66<X<71) = 0.4679

The probability that the height is between 66 and 71 inches is 0.4679.

(c) To find the probability P(X>71), we need to find P(X<71) and then subtract it from 1 because the normal distribution table gives values for P(X<k). We have already calculated the value of P(X<71) in part (b) so,

P(X>71) = 1 - P(X<71)

= 1 - 0.7422

P(X>71) = 0.2578

The probability that the height is more than 71 inches is 0.2578.

3 0

3 years ago

Need help please worth 50 points

Law Incorporation [45]

Here's how you simplify those. The rule for raising an exponent to an exponent is that you multiply them. The rule for dividing exponents with the same base is that you subtract the denomonator from the numerator. So let's simplify the first one, which is actually the most confusing.
$( \frac{ p^{5} }{ p^{-3} q^{-4} } )^{ \frac{1}{4} }$
Multiplying the exponents you get:
$( \frac{ p^{ \frac{5}{4} } }{ p^{- \frac{3}{4} } q^{-1} })$
Subtracting the denominator from the numerator between the common base of p you get this:
$\frac{ p^{ \frac{5}{4}- (-\frac{3}{4}) } }{ q^{-1} }$
Doing that math gives you
$\frac{ p^{2} }{ \frac{1}{q} }$ which equals $\frac{ p^{2} }{1} * \frac{q}{1}$ which is $p^{2}q$ , or the third one down.

The next one:
$( \frac{ p^{2} q^{7} }{ q^{4} } )^{ \frac{1}{2} }$
simplifies to:
$\frac{p* q^{ \frac{7}{2} } }{ q^{2} }$
and subtracting the power of the denominator from the power of the numerator gives you:
$p q^{ \frac{3}{2} }$ , which is the last choice.
$( p^{6}q \frac{3}{2})^{ \frac{1}{3} }$
simplifies to:
$( p^{6* \frac{1}{3} } )( q^{ \frac{3}{2}* \frac{1}{3} })$
which simplifies very nicely to:
$p^{2}q^{ \frac{1}{2} }$ , which is the first choice. The other one is found by process of elimination!

8 0

4 years ago

Answer for a lot of points!

earnstyle [38]

Given :

ZC = 90°

CD is the altitude to AB.

$\angle$ A = 65°.

To find :

the angles in △CBD and △CAD if m∠A = 65°

Solution :

In Right angle △ABC,

we have,

=> ACB = 90°

=> $\angle$ CAB = 65°.

So,

=> $\angle$ ACB + $\angle$ CAB+ $\angle$ ZCBA = 180° (By angle sum Property.)

=> 90° + 65° + $\angle$ CBA = 180°

=> 155° + $\angle$ CBA = 180°

=> $\angle$ CBA = 180° - 155°

=> $\angle$ CBA = 25°.

In △CDB,

=> CD is the altitude to AB.

So,

=> $\angle$ CDB = 90°

=> $\angle$ CBD = $\angle$ CBA = 25°.

So,

=> $\angle$ CBD + $\angle$ DCB = 180° (Angle sum Property.)

=> 90° +25° + $\angle$ DCB = 180°

=> 115° + $\angle$ DCB = 180°

=> $\angle$ DCB = 180° - 115°

=> $\angle$ DCB = 65°.

Now, in △ADC,

=> CD is the altitude to AB.

So,

=> $\angle$ ADC = 90°

=> $\angle$ CAD = $\angle$ CAB = 65°.

So,

=> $\angle$ ADC + $\angle$ CAD + $\angle$ DCA = 180° (Angle sum Property.)

=> 90° + 65° + $\angle$ DCA = 180°

=> 155° + $\angle$ DCA = 180°

=> $\angle$ DCA = 180° - 155°

=> $\angle$ DCA = 25°

Hence, we get,

$\angle$ DCA = 25°
$\angle$ DCB = 65°
$\angle$ CDB = 90°
$\angle$ ACD = 25°
$\angle$ ADC = 90°.

7 0

3 years ago

Think about a standard deck of 52 playing cards. Which of the following events is mutually exclusive?

Shtirlitz [24]

Answer:

1.) Drawing a king or a queen.

Step-by-step explanation:

1.) Drawing a king or a queen.

You cannot get a king and a queen at the same time= mutually exclusive

2.) Drawing a club or a black card.

King of clubs is both a club and a black card

3.) Drawing a king or a heart.

king of hearts is both a heart and a king

4.) Drawing a face card or a heart

queen of hearts is a heart and a face card

7 0

3 years ago

Please help I will give Brainliest please!

WITCHER [35]

Part (a)

The domain is the set of allowed x inputs of a function.

The graph shows that x = 0 is not allowed because of the vertical asymptote located here. It seems like any other x value is fine though.

<h3>Domain: set of all real numbers but $x \ne 0$ </h3>

To write this in interval notation, we can say $(-\infty, 0) \cup (0, \infty)$ which is the result of poking a hole at 0 on the real number line.

--------------

The range deals with the y values. The graph makes it seem like it stretches on forever in both up and down directions. If this is the case, then the range is the set of all real numbers.

<h3>Range: Set of all real numbers</h3>

In interval notation, we would say $(-\infty, \infty)$ which is almost identical to the interval notation of the domain, except this time of course we aren't poking at hole at 0.

=======================================================

Part (b)

<h3>The x intercepts are x = -4 and x = 4</h3>

We can compact that to the notation $x = \pm 4$

These are the locations where the blue hyperbolic curve crosses the x axis.

=======================================================

Part (c)

<h3>Answer: There aren't any horizontal asymptotes in this graph.</h3>

Reason: The presence of an oblique asymptote cancels out any potential for a horizontal asymptote.

=======================================================

Part (d)

The vertical asymptote is located at x = 0, so the equation of the vertical asymptote is naturally x = 0. Every point on the vertical dashed line has an x coordinate of zero. The y coordinate can be anything you want.

<h3>Answer: x = 0 is the vertical asymptote</h3>

=======================================================

Part (e)

The oblique or slant asymptote is the diagonal dashed line.

It goes through (0,0) and (2,6)

The equation of the line through those points is y = 3x

If you were to zoom out on the graph (if possible), then you should notice the branches of the hyperbola stretch forever upward but they slowly should approach the "fencing" that is y = 3x. The same goes for the vertical asymptote as well of course.

<h3>Answer: Oblique asymptote is y = 3x</h3>

5 0

3 years ago