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Travka [436]
3 years ago
10

Use the balanced scale to find the conversion factor that can be used to convert the number of blocks to the weight of the block

s, in pounds?
Mathematics
2 answers:
adoni [48]3 years ago
5 0

Answer:

2.5

Step-by-step explanation:

since there is six 1 lb blocks, it is 2.5 because 6 lb to 2 blocks will be 2.5 because its half of 6

marusya05 [52]3 years ago
4 0

Answer:

the answer is 24 lb

i think

Stepkby-step explanation:

it is 24 because it is 6 1lb blocks x 4 will give you 24

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A coin is tossed twice. What is the probability of getting a tail in the first toss and a tail in the second toss?
skelet666 [1.2K]

Answer:

<h2>1/4 Chances</h2><h2>25% Chances</h2><h2>0.25 Chances (out of 1)</h2>

Step-by-step explanation:

Two methods to answer the question.

Here are presented to show the advantage in using the product rule given above.

<h2>Method 1:Using the sample space</h2>

The sample space S of the experiment of tossing a coin twice is given by the tree diagram shown below

The first toss gives two possible outcomes: T or H ( in blue)

The second toss gives two possible outcomes: T or H (in red)

From the three diagrams, we can deduce the sample space S set as follows

          S={(H,H),(H,T),(T,H),(T,T)}

with n(S)=4 where n(S) is the number of elements in the set S

tree diagram in tossing a coin twice

The event E : " tossing a coin twice and getting two tails " as a set is given by

          E={(T,T)}

with n(E)=1 where n(E) is the number of elements in the set E

Use the classical probability formula to find P(E) as:

          P(E)=n(E)n(S)=14

<h2>Method 2: Use the product rule of two independent event</h2>

Event E " tossing a coin twice and getting a tail in each toss " may be considered as two events

Event A " toss a coin once and get a tail " and event B "toss the coin a second time and get a tail "

with the probabilities of each event A and B given by

          P(A)=12 and P(B)=12

Event E occurring may now be considered as events A and B occurring. Events A and B are independent and therefore the product rule may be used as follows

        P(E)=P(A and B)=P(A∩B)=P(A)⋅P(B)=12⋅12=14

NOTE If you toss a coin a large number of times, the sample space will have a large number of elements and therefore method 2 is much more practical to use than method 1 where you have a large number of outcomes.

We now present more examples and questions on how the product rule of independent events is used to solve probability questions.

8 0
3 years ago
Read 2 more answers
PLEASE HELP!!!
viktelen [127]

The answer is 3/5.


\frac{1}{3} *\frac{3}{5} =\frac{3}{15} \\ \\ \frac{3}{15} *\frac{3}{5} =\frac{9}{75}


Continue this pattern and you'll see that it follows this geometric sequence

5 0
3 years ago
Read 2 more answers
Math help what is the answer will give brainliest.
artcher [175]
The two equations are
x + y = 7
and
4x + 2.5y = 22
the first equation can be solved for x to be
x = 7 - y
next you can substitute x in the second equation
4(7 - y) + 2.5y = 22
Distribute 4 to everything in the parentheses
28 - 4y + 2.5y = 22
combine like terms:
28 - 1.5y = 22
subtract 28 from both sides of the equation
- 1.5y =  - 6
divide both sides by -1.5
y =  4
plug this in to the original equation of x+y=7
x + 4 = 7
subtract 4 from both sides.
x = 3
Final Answer: they bought 3 adult tickets and 4 student tickets


Hope this helps!
7 0
3 years ago
HELP PLEASE!!! Find the values of x and y. Express your answers in simplest radical form.
Alchen [17]
Answer:
C.

Explanation: I guest it is base on my research but i dont have a solution peace
7 0
3 years ago
WILL GIVE BRAINLIEST
Sindrei [870]
B jxkdjxjzjzkskslsmxndd she k
4 0
3 years ago
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