1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Colt1911 [192]
3 years ago
9

HELP ASAP!! Please will give branliest if correct! :)

Mathematics
1 answer:
Kitty [74]3 years ago
8 0
Look at each of the points. Specifically note the y coordinate of each point.

The points are (1,4), (2,6), (3,8) and (4,10)
The y values of those points are: 4, 6, 8, 10

Therefore, the sequence is: 4, 6, 8, 10
The values will go in the boxes in that order
You might be interested in
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
Angle Relationships
Snowcat [4.5K]

Answer:

B.

   BC = 4.8cm

   AC = 3.6cm

   ∠B = 37°

   ∠C =90°

Step-by-step explanation:

I calculated it logically

4 0
2 years ago
704 rounded to the nearest ten
Assoli18 [71]
The answer is 700 because the 4 rounds down.
5 0
3 years ago
Read 2 more answers
Write an absolute value inequality that has 3 and -5 as two of its solutions
tino4ka555 [31]
Absolute value makes all things inside positive (make result positive)
so one example is
|x+1|=4
so if you had 3 in it woul dyhave
|3+1|=4
|4|=4
4=4
true

if -5
|-5+1|=4
|-4|=4
4=4
treu
8 0
3 years ago
If you have 10 boxes and 93 needs to be distributed in to them equally how much is in each box
dalvyx [7]
9.3 due to the fact that 93/10 is 9.3 hopefully that helps
7 0
3 years ago
Read 2 more answers
Other questions:
  • Two brothers shared the coins in a piggy bank. Alan received 140 coins. When he gave 48 coins to his brother, the number of his
    13·1 answer
  • Solve for x 2x=-16 x=
    5·2 answers
  • In the figure above, lines l and m are parallel and
    14·1 answer
  • Can someone please help me find the domain, range, intercepts, and asymptotes pf this function?
    5·1 answer
  • Help and explain if you can please ​
    11·1 answer
  • What is the following quotient 6-3(^3+6)/^379
    10·1 answer
  • Which statements must be true about the reflection of
    14·2 answers
  • Question 11 (5 points)
    13·1 answer
  • The Rodriguez family took 5.5 hours to gather information for their Schedule A. The itemized deductions saved them $4,095 in tax
    6·1 answer
  • HELP <br> Write an equation of the line in slope intercept form using the graph below.
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!