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viva [34]
3 years ago
14

Jack was 99 years older than priscilla. together their ages totaled 167167 years. what were their​ ages?

Mathematics
1 answer:
faust18 [17]3 years ago
6 0
Priscilla's age was 83,534 while Jack's age was 83,633.
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A father and his son decide to sum their age. The sum is equal to sixty years. Six years ago, the age of the father was five tim
arlik [135]

Answer:

20

Step-by-step explanation:

To solve this you have to make a system of equations.

Since the father and son's age sum up to 60, the first equation will be:

f + s = 60

Secondly, since the father's age is 5 times the age of the son 6 years ago the equation will be:

6 - (5s) = f

Now, you have to solve the first equation to let it equal to s

f + s = 60

f = 60 - s

Plug in

6 - 5s = 60 - s

     +s          +s

6 - 4s = 60

-6         -6

---------------------

-4s = 54

-----   -----

 -4     -4

   s ≅ 14

14 + 6 = 20

5 0
1 year ago
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Rosa is putting cube-shaped tissue boxes into a shipping crate. each tissue box has a side length of 4 inches. The volume of V o
Elena L [17]

Answer:

V = 64 and V = 2,048

Step-by-step explanation:

The volume of a cube is found using V = s*s*s. Substitute s = 4.

The volume is V = 4*4*4 = 64.

The volume of the crate is 32 * 64 = 2,048.

3 0
3 years ago
What is the general form of the equation of the line shown?
Arturiano [62]
The answer is A, because the Y-intersect starts at -2
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A tax of 20% is collected on every purchase at a supermarket.
mafiozo [28]
The first three statements are correct while the last statement is incorrect.
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3 years ago
Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of
IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  • np ≥ 10
  • n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

8 0
3 years ago
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