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morpeh [17]
3 years ago
14

sue and 4 friends buy a box of 362 base ball cards at a yard sale. If they share the cards equally,how many cards will each pers

on receive
Mathematics
1 answer:
Radda [10]3 years ago
4 0
They should receive 72 each 362 divided by 5 (number people)
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At a certain middle school, there are 26 students per teacher in every homeroom. Is the total number of students proportional to
Verizon [17]

Yes, because there is a corresponding amount of teachers per student.


6 0
3 years ago
Please review the following attachment. Thanks!
vaieri [72.5K]

Answer:

C: I and III only.

Step-by-step explanation:

We have two numbers x and y such that:

-y

And we want to determine which of the following conditions must be true.

First, let's examine the third condition. We have:

y>0

To see if this is true, let's try a negative value for y. So, let's use -7. This will give:

-(-7)

Simplify:

7

This is saying we need a number less than -7 and greater than 7.

This is impossible. So, y <em>must be</em> greater than 0.

Therefore, Condition III must be true.

Next, let's see this compound inequality visually.

Picture the following number-line:

<----------(-y)----------0----------y---------->

Whatever y is, -y is just y on the negative side.

Now, since our condition is that -y<x<y, this means that our x can be anywhere between -y and y. Namely:

<----------(-y)----------0----------y---------->

To determine our correct condition, let's picture x anywhere on the bolded lines. I'm just going to put x here...

<----------(-y)-------(x)--0----------y---------->

Now, remember the alternative definition for absolute value. Namely, the absolute value of x is also the <em>distance</em> from 0 to x. And this distance is always positive.

Since our x is between -y and y, our distance from 0 to x will always be less than the distance from 0 to either y.

Therefore, Condition I is also true.

Let's try an example. Let's let y = 9. So, -y=-9. And let's let x be between 9 and -9, say, -2. So:

<---(-9)-----(-2)--0--------(9)------>

We can see that:

|-2|=2\stackrel{\checkmark}{

Also, this example counters Condition II, as our x can indeed be negative if we desire it.

Algebraically, if we have a negative x such that:  

-y

We can divide everything by -1 to obtain:

y>x>-y

We can flip this to get:

-y

Which is our original inequality. So, Condition II does not need to be true.

So, the two conditions that <em>must</em> be true is I and III.

So, our answer is C.

3 0
3 years ago
Please help me out with this one! I need step by step instructions (30 points!)
jolli1 [7]

Answer:

WAT DO U MEAN 30 POINTS UR ONLY GIVING 5

Step-by-step explanation:

ITS LITTERALLY RIGHT NEXT TO IT

8 0
3 years ago
The ratio of shirts to pants is 3 to 2. if there are 6 shirts, how many pants are there?
marshall27 [118]
4 pants.
 
I hope this helps, please Brainliest me, and have a great day! =D

3 0
3 years ago
Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the
Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million)     Number of companies

25 up to 35                                                    4

35 up to 45                                                    19

45 up to 55                                                    27

55 up to 65                                                    16

65 up to 75                                                     9

TOTAL                                                            75

Let's find the probabilities first:

P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}

For 35 up to 45

P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}

For 45 up to 55

P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}

For 55 up to 65

P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}

For 65 up to 75

P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}

Chi-Square Table can be computed as follows:

Expense   No of   Probabilities(P)  Expe                (O-E)^2   \dfrac{(O-E)^2}{E}

             compa                                 cted E (n*p)

             nies (O)  

25-35           4      0.0612    75*0.0612 = 4.59        0.3481       0.0758

35-45           19     0.2218   75*0.2218 = 16.635     5.5932       0.3362

45-55           27     0.3568   75*0.3568 = 26.76     0.0576      0.021

55-65           16      0.2552   75*0.2552 = 19.14      9.8596      0.5151

65-75           9      0.0839     75*0.0839 = 6.2925   7.331         1.1650

                                                                                           \sum \dfrac{(O-E)^2}{E}= 2.0492                                                                                                      

Using the Chi-square formula:

X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square  \ X^2 = 2.0942

Null hypothesis:

H_o: \text{The population of advertising expenses follows a normal distribution}

Alternative hypothesis:  

H_a: \text{The population of advertising expenses does not follows a normal distribution}

Assume that:

\alpha = 0.02

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from X^2 = 11.667

Decision rule: To reject H_o  \  if \  X^2  test statistics is greater than X^2 tabulated.

Conclusion: Since X^2 = 2.0942 is less than critical value 11.667. Then we fail to reject H_o

6 0
3 years ago
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