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vladimir1956 [14]
4 years ago
5

the ratio of the number of Annes pencils to the number of jasons pencils is 4:3. Anne has 100 pencils how many pencils does Jaso

n have
Mathematics
1 answer:
S_A_V [24]4 years ago
4 0

Answer:

Jason has 75 pencils.

Step-by-step explanation:

In order to get from 4 to 100, you must multiply by 25 (100/4 = 25). And in order to find the amount of pencils Jason has, you must have an equivalent ratio. So you have to also multiply 3 by 25, 3 * 25 = 75. Jason has 75 pencils.

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OLEGan [10]

Answer:

I believe it would be 101,000

Step-by-step explanation:

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There were 125 females and 100 males present at the high school pep rally. Find the ratio of males to the total number of people
riadik2000 [5.3K]
Total number of people = 125 + 100 = 225

Number of males = 100

Set as ratio

100:225

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3 years ago
What is the length of x in the triangle shown
ivann1987 [24]
Since this is a right triangle the angle opposite the side 17 is also 45°

So this can be solved in two ways...

The Law of sines...

x/sin90=17/sin45

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8 0
3 years ago
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8 0
3 years ago
Each of a sample of four home mortgages is classified as fixed rate (F) or variable rate (V). (Enter your answers as a comma-sep
omeli [17]

Answer:

a)

{FVVV,FFVV,VFVV,VVVV,FVFV,FFFV,VFFV,VVFV,FVVF,FFVF,VFVF,VVVF,FVFF,FFFF,VFFF,VVFF}

b)

{FFFV,FFVF,FVFF,VFFF}

c)

{FFFF,VVVV}

d)

{FFFV,FFVF,FVFF,FFFF,VFFF}

e)

{FFFV,FFVF,FVFF,FFFF,VFFF,VVVV}

{FFFF}

f)

{FFFV,FFVF,FVFF,VFFF,FFFF,VVVV}

{ }

Step-by-step explanation:

a)

The 16 outcomes for a sample of four home mortgages classified as fixed and variable rate are

Sr.No Outcomes

1          FVVV

2         FFVV

3         VFVV

4         VVVV

5         FVFV

6         FFFV

7         VFFV

8         VVFV

9         FVVF

10        FFVF

11         VFVF

12        VVVF

13        FVFF

14        FFFF

15        VFFF

16        VVFF

b)

Let A be the event that exactly three are fixed rate. Thus, event A consists of following outcomes

A={FFFV,FFVF,FVFF,VFFF}

c)

Let B be the event that all four are of same type. The event B consists of either all fixed rate or all variable rate. Thus, event B consists of following outcomes

B={FFFF,VVVV}

d)

Let C be the event that at most one of four is variable rate. The event C consists of less than and equal to one variable rate. Thus, event C consists of following outcomes

C={FFFV,FFVF,FVFF,FFFF,VFFF}

e)

Let union of part(c) and part(d) can be represented as BUC. The union represents all the outcomes in event B and event C.

B∪C={FFFF,VVVV}∪{FFFV,FFVF,FVFF,FFFF,VFFF}

B∪C={FFFV,FFVF,FVFF,FFFF,VFFF,VVVV}

Let intersection of part(c) and part(d) can be represented as B∩C. The intersection represents the common outcomes in event B and event C.

B∩C={FFFF,VVVV}∩{FFFV,FFVF,FVFF,FFFF,VFFF}

B∩C={FFFF}

f)

Let union of part(b) and part(c) can be represented as AUB. The union represents all the outcomes in event A and event B.

A∪B={FFFV,FFVF,FVFF,VFFF}∪{FFFF,VVVV}

A∪B={FFFV,FFVF,FVFF,VFFF,FFFF,VVVV}

Let intersection of part(b) and part(c) can be represented as A∩B. The intersection represents the common outcomes in event A and event B.

A∩B={FFFV,FFVF,FVFF,VFFF}∩{FFFF,VVVV}

A∩B={ }

So intersection of part(b) and part(c) is an empty set as there is no common outcome in these two sets.

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