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3 years ago

8

Joey spins a spinner and lands on the number 4. Then Jennifer spins and lands on the number 2. Are these events independent or d

ependent?

1 answer:

miv72 [106K]3 years ago

7 0

Answer:

Independent. The spins do not have any connections nor do the players.

Step-by-step explanation:

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Find the area of the circle. Round your answer to the nearest tenth. The circle has a radius of 9

Rudik [331]

Answer:

I is at least ten because of nine it rounds to ten

3 0

3 years ago

What is the slope of the line with equation 6x+3y=12?

max2010maxim [7]

Answer:

<h2>The slope m = -2</h2>

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

<em>m</em><em> - slope</em>

<em>b</em><em> - y-intercept</em>

<em />

We have the equation in the standard form:

$6x+3y=12$

Convert to the slope intercept form:

$6x+3y=12$ <em>subtract 6x from both sides</em>

$3y=-6x+12$ <em>divide both sides by 3</em>

$y=-2x+4$

slope: <em>m = -2</em>

y-intercept: <em>b = 4</em>

6 0

3 years ago

The given box plots show the number of text messages Paul and Sally received each day on their cell phones.

zavuch27 [327]

To be able to answer this we need to see the box plots and the rest of the question. Sorry -♡

7 0

3 years ago

A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)

Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

$36^6$

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

$5\cdot 36^5$

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

$\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}$

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

$P(A)=P(B)=\dfrac{5}{36}$

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

$axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8$

$exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8$

$ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8$

$oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8$

$uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8$

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

$P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}$

Finally, you can compute the probability of the union using the formula

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Since we already computed all these quantities.

7 0

2 years ago

Item 6 for what value of a is 8x−8+3ax=5ax−2a an identity?

Readme [11.4K]

Let's rewrite the expression as

$8x+3ax-5ax = 8-2a \iff 8x - 2ax = 8-2a \iff 2x(4-a) = 2(4-a)$

So, if $a \neq 4$ , you may divide both sides by $4-a$ , the equation becomes $2x=2$ , and the only solution is $x=1$ , which means that this is not an identity (an indentity is tautologically true, no matter the value of x).

If instead $a = 4$ , you are multiplying both sides by zero, so the equation becomes

$2x \cdot 0 = 2 \cdot 0 \iff 0=0$

So, it doesn't matter which value for x you choose, because you will always end up with $0=0$ , which is obviously always true.

8 0

3 years ago