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Ludmilka [50]
2 years ago
12

The spinner below is spun twice. If the spinner lands on a border, that spin does not count and spin again. It is equally likely

that the spinner will land in each of the six sectors.
RED
RED
RED
BLUE
BLUE
CYAN
[Graphs generated by this script: initPicture(-100,100,-100,100);stroke='red'; fill='red';arc([90,0],[45,77.94228],90);path([[0,0],[90,0],[45,77.94228],[0,0]]);arc([-45,77.94228],[-90,0],90);path([[0,0],[-90,0],[-45,77.94228],[0,0]]);arc([-45,-77.94228],[45,-77.94228],90);path([[0,0],[-45,-77.94228],[45,-77.94228],[0,0]]);stroke='blue'; fill='blue';arc([45,77.94228],[-45,77.94228],90);path([[0,0],[45,77.94228],[-45,77.94228],[0,0]]);arc([-90,0],[-45,-77.94228],90);path([[0,0],[-90,0],[-45,-77.94228],[0,0]]);stroke='cyan'; fill='cyan';arc([45,-77.94228],[90,0],90);path([[0,0],[45,-77.94228],[90,0],[0,0]]);stroke='black'; fill='none';circle([0,0],90);line([45,77.94228],[-45,-77.94228]);line([-45,77.94228],[45,-77.94228]);line([-90,0],[90,0]);fontfamily='helvetica';fontfill='white';fontstyle='normal';fontweight='bold';text([55,25],'RED');text([-50,25],'RED');text([0,-70],'RED');text([0,65],'BLUE');text([-55,-30],'BLUE');fontfill='black';text([50,-25],'CYAN');strokewidth='5';line([-35,-35],[30,30]);line([25,15],[30,30]);line([15,25],[30,30]);]

For each question below, enter your response as a reduced fraction.

Find the probability of spinning blue on the first spin and cyan on the second spin.


Find the probability of spinning red on the first spin and cyan on the second spin.


Find the probability of NOT spinning red on either spin. (Not red on the first spin and not red on the second spin.)
Mathematics
1 answer:
cestrela7 [59]2 years ago
3 0

Answer:

A) 1/6 & 1/12 (with borders) 1/3 & 1/6 (without borders)

B) 1/4 & 1/12 (with borders) 1/2 & 1/6 (without borders)

C) 1/4 & 1/4  (with borders) 1/2 & 1/2 (without borders)

Step-by-step explanation:

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If Conny had x apples and Johnny had 56
ycow [4]

x+56=103

Subtract 56 from both sides.

x=103-56

x=47

x is the number of apples Conny has, so Conny has 47 apples.

I hope this helps :)

6 0
3 years ago
Read 2 more answers
Frankie earns $5 each time he babysits his little sister. He has saved $30. Frankie wants to save $52 to buy a new skateboard. W
inessss [21]

Answer:

Frankie will need to babysit for 4.4 time.

Step-by-step explanation:

We are given the following in the question:

Money earned by Frankie each times he babysits = $5

Money saved = $30

Total money Frankie wants to save for a skateboard = $52

Amount of money to be saved =

Total amount of money Frankie wants to save for a skateboard - Money saved already

= 52-30\\=22\$

He needs to save $22 more.

Let x be the number of times Frankie will need to babysit.

Thus, we can write the equation:

5x = 22\\\\\Rightarrow x = \dfrac{22}{5}\\\\\Rightarrow x = 4.4

Thus, Frankie will need to babysit for 4.4 time.

7 0
3 years ago
A restaurant offers
Neporo4naja [7]
63 two-course meals can be made.
7 0
4 years ago
Millie has 5 yards of blue fabric and 7 yards of pink fabric. How many quilt squares can she make with the fabric she has if bot
inn [45]
Divide them and you're answer will come out
7 0
3 years ago
Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
3 years ago
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