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
otez555 [7]
3 years ago
5

If the date of the second Thursday of the month is a perfect square, what is the date of the last Monday of the month?

Mathematics
1 answer:
Inga [223]3 years ago
3 0

Answer:

last Monday is 27

Step-by-step explanation:

perfect square: 4, 9 , 16

2nd Thursday is a day >5 and < 14 day from the begining of the month, therefore the only day fit the criteria is 9th of the month.

last Monday is 27

Sun Mon Tue Wen Thu Fri Sat

                                  9     10  11

12    13      14     15     16     17  18

.........................................................

26    27

You might be interested in
Gia bought a pound of cheese and used 1/4 of it. She split the rest into 5 equal portions for lunch during the week. Which expre
Reptile [31]
Gia already used 1/4. 3/4 divided by 5 = 3/20
7 0
3 years ago
Brice has $1200 in the bank. He wants to save a total of $3000 by depositing $40 per week from his paycheck. Write and use an eq
Anon25 [30]

$$3000-$1200=$1800

take $1800÷$40=45

<h3>answ=45 weeks</h3>

6 0
2 years ago
A coin is biased such that it results in 2 heads out of every 3 coins flips on average
alina1380 [7]

<span>The mathematical theory of probability assumes that we have a well defined repeatable (in principle) experiment, which has as its outcome a set of well defined, mutually exclusive, events.</span>


If we assume that each individual coin is equally likely to come up heads or tails, then each of the above 16 outcomes to 4 flips is equally likely. Each occurs a fraction one out of 16 times, or each has a probability of 1/16.

Alternatively, we could argue that the 1st coin has probability 1/2 to come up heads or tails, the 2nd coin has probability 1/2 to come up heads or tails, and so on for the 3rd and 4th coins, so that the probability for any one particular sequence of heads and tails is just (1/2)x(1/2)x(1/2)x(1/2)=(1/16).

Now lets ask: what is the probability that in 4 flips, one gets N heads, where N=0, 1, 2, 3, or 4. We can get this just by counting the number of outcomes above which have the desired number of heads, and dividing by the total number of possible outcomes, 16. 
  
 

<span>N     # outcomes with N heads     probability to get N heads</span>

0                1                                       1/16 = 0.0625

1                4                                       4/16 = 1/4 = 0.25

2                6                                      6/16 = 3/8 = 0.375

3                4                                      4/16 = 1/4 = 0.25

4                1                                      1/16 = 0.0625

We can plot these results on a graph as shown below.

 
The dashed line is shown just as a guide to the eye. Notice that the curve has a "bell" shape. The most likely outcome is for N=2 heads, where the curve reaches its maximum value. This is just what you would expect: if each coin is equally likely to land heads as tails, in four flips, half should come up heads, that is N = 4x(1/2) = 2 is the most likely outcome. Note however that an occurrence of N = 1 or N = 3 is not so unlikely - they occur 1/4 or 25% of the time. To have an occurrence of only N = 0, or N = 4 (no heads, or all heads) is much less likely - they occur only 1/16 or 6.25% of the time.

The above procedure is in principle the way to solve all problems in probability. Define the experiment, enumerate all possible mutually exclusive outcomes (which are usually assumed to be each equally likely), and then count the number of these outcomes which have the particular property being tested for (here for example, the number of heads). Dividing this number by the total number of possible outcomes then gives the probability of the system to have that particular property.

Often, however, the number of possible outcomes may be so large that an explicit enumeration would become very tedious. In such cases, one can resort to more subtle thinking to arrive at the desired probabilities. For example, we can deduce the probabilities to get N heads in 4 flips as follows:

N=0: There is only one possible outcome that gives 0 heads, namely when each flip results in a tail. The probability is therefore 1/16.

N=4: There is only one possible outcome that gives 4 heads, namely when each flip results in a head. The probability is therefore 1/16.

N=1: There are 4 possible outcomes which will have only one coin heads. It may be that the 1st coin is heads, and all others are tails; or it may be that the 2nd coin is heads, and all others are tails; or it may be that the 3rd (or the 4th) coin is heads, and all others are tails. Since there are 4 possible outcomes with one head only, the probability is 4/16 = 1/4.

N=3: To get 3 heads, means that one gets only one tail. This tail can be either the 1st coin, the 2nd coin, the 3rd, or the 4th coin. Thus there are only 4 outcomes which have three heads. The probability is 4/16 = 1/4.

N=2: To enumerate directly all the possible outcomes which have exactly 2 heads only, is a bit trickier than the other cases. We will come to it shortly. But we can get the desired probability for N=2 the following way: We have already enumerated all possible outcomes with either N = 0, 1, 3, or 4 heads. These account for 1 + 4 + 4 + 1 = 10 possible outcomes. The only outcomes not include in these 10 are those with exactly N=2 heads. Since there are 16 possible outcomes, and 10 do not have N=2 heads, there must therefore be exactly 16 - 10 = 6 outcomes which do have exactly N=2 heads. The probability for N=2 is therefore 6/16 = 3/8.

2) Consider the experiment of rolling 3 dice, each of which has 6 sides.

What is the probability that no two dice land with the same number side up, i.e. each of the three dice rolls a different number?

Since each die has 6 possible outcomes, the number of possible outcomes for the roll of three dice is 6x6x6 = 216. We could enumerate all these 216 possibilities, and then count the number of outcomes in which each die has a different number. This is clearly too tedious! Instead we reason as follows:


6 0
2 years ago
Read 2 more answers
Figure AAA is a scale image of figure BBB.
Elan Coil [88]

Answer:

4

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
PLZ SOLVE ASAP Solve for f. -11f=7(1-2f)+5
viktelen [127]

Answer:

f = 4

Step-by-step explanation:

Step 1: Distribute

-11f = 7 - 14f + 5

Step 2: Combine like terms

-11f = 12 - 14f

Step 3: Add 14f to both sides

3f = 12

Step 4: Divide both sides by 3

f = 4

4 0
3 years ago
Read 2 more answers
Other questions:
  • What is 751,447 rounded to the nearest hundred thousand?
    9·2 answers
  • Represent the above relationship between the number of triangles and the perimeter of the figures they form by filling in the ta
    13·2 answers
  • Help I need somebody help
    14·1 answer
  • Two different floor plans are being offered in a new housing development. Several prospective
    12·1 answer
  • 1.5 cups per day how many gallons per weeek? (1gallon =16cup)
    15·1 answer
  • An mp3 player is on sale for 25% off. The original price is $52.60. What is the sale price?
    12·1 answer
  • Ariana has a collection of 100 coins. How many coins represent 10% of her collection?
    7·1 answer
  • Why are the triangles congruent? (Need ASAP 15points)
    14·2 answers
  • Please help me in this!!!
    10·1 answer
  • this is a question on a review assignment that i’m stuck on , my summer school teacher said i could get help if i needed. any he
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!