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Bezzdna [24]
3 years ago
8

Which number is not a common factor of 32 and 48

Mathematics
1 answer:
Colt1911 [192]3 years ago
8 0
5, 3, 7, 9 are all non-common factors
You might be interested in
For which positive integer values of $k$ does $kx^2+20x+k=0$ have rational solutions? Express your answers separated by commas a
Musya8 [376]

Answer:

-10,10

Step-by-step explanation:

The given quadratic equation is

k {x}^{2}  + 20x + k = 0

The discriminant of this equation is given by;

D =  {b}^{2}  - 4ac

where a=k, b=20, c=k

For rational solutions, the discriminant must be zero.

{20}^{2}  - 4 \times k \times k = 0

Simplify to get:

400 - 4  {k}^{2}  = 0

This implies that:

400  = 4  {k}^{2}

100 =  {k}^{2}

Take square root to get:

k =   \pm\sqrt{100}

k =  \pm10

k =  - 10 \: or \: k = 10

3 0
3 years ago
You roll a pair of honest dice. If you roll a total of 7, you win $18; if you roll a total of 11, you win $54; if you roll any o
amm1812
Answer: -1
The negative value indicates a loss

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

Explanation:

Define the three events
A = rolling a 7
B = rolling an 11
C = roll any other total (don't roll 7, don't roll 11)

There are 6 ways to roll a 7. They are
1+6 = 7
2+5 = 7
3+4 = 7
4+3 = 7
5+2 = 7
6+1 = 7
Use this to compute the probability of rolling a 7
P(A) = (number of ways to roll 7)/(number total rolls) = 6/36 = 1/6
Note: the 36 comes from 6*6 = 36 since there are 6 sides per die

There are only 2 ways to roll an 11. Those 2 ways are:
5+6 = 11
6+5 = 11
The probability for event B is P(B) = 2/36 = 1/18

Since there are 6 ways to roll a "7" and 2 ways to roll "11", there are 6+2 = 8 ways to roll either event. 
This leaves 36-8 = 28 ways to roll anything else
P(C) = 28/36 = 7/9

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

In summary so far,
P(A) = 1/6
P(B) = 1/18
P(C) = 7/9

The winnings for each event, let's call it W(X), represents the prize amounts.
Any losses are negative values
W(A) = amount of winnings if event A happens 
W(B) = amount of winnings if event B happens
W(C) = amount of winnings if event C happens
W(A) = 18
W(B) = 54
W(C) = -9

Multiply the probability P(X) values with the corresponding W(X) values
P(A)*W(A) = (1/6)*(18) = 3
P(B)*W(B) = (1/18)*(54) = 3
P(C)*W(C) = (7/9)*(-9) = -7

Add up those results
3+3+(-7) = -1

The expected value for this game is -1.
The player is expected to lose on average 1 dollar per game played.


Note: because the expected value is not 0, this is not a fair game.


7 0
3 years ago
Read 2 more answers
Help me please!<br> What is the median of this data? 48, 60, 22, 84, 36
vova2212 [387]
He is wrong. You first need to put it in order before getting the middle. Like so:
22, 36, 48, 60, 84
Now cross out 22, then 84, and so on till you're left with one number (the middle number)

THE ANSWER IS 48
3 0
3 years ago
Read 2 more answers
The mean exam score for the first group of twenty examinees applying for a security job is 35.3 with a standard deviation of 3.6
babymother [125]

The scores of two groups can be compared using coefficient of variation;

Coefficient of variation (C.V.) = (Standard Deviation/ Mean) × 100%;

For Data set 1;

Standard deviation = 3.6

Mean = 35.3

C.V. = (3.6/35.3) × 100%;

       = 10.19%

For Data set 2;

Standard deviation = 0.5

Mean = 34.1

C.V. = (0.5/34.1) × 100%;

       = 1.46%

To learn more about coefficient of variation, visit: brainly.com/question/24131744

#SPJ9

6 0
2 years ago
for a field trip 12 students rode in cars and the rest filled six buses how many students were in each bus if 342 students were
kolezko [41]

Answer:

55

Step-by-step explanation:

we have 342 students, subtract 12 because we need only kids riding the bus, now we have 330 kids, we divide the 330 kids between 6 buses 330/6, our answer is 55.

4 0
3 years ago
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