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natita [175]
2 years ago
13

There are 5 cars and 7 trucks in a parking lot. James says that 50% are cars . do you agree or disagree with James

Mathematics
1 answer:
Leno4ka [110]2 years ago
4 0
Disagree because there’s more trucks than cars so half can’t be cars
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A pack of red pens can be divided into equal shares among 3, 5, 6 and 9 teachers with no pen left over?
Ugo [173]

Answer:

90 red pens

Step-by-step explanation:

we need to find the least common multiple of the numbers 3 , 5 , 6 , 9

6 = 2 × 3

9 = 3²

then

LCM(3,5,6,9) = 2×5×3² = 2×5×9 = 90

The number 90 can be divided into 3 ,5 ,6 ,9 with no remainder

Therefore the pack must have 90 pens or a multiple of 90

For example 180 , 270 ...

7 0
2 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
2 years ago
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7nadin3 [17]
I would say 7/24 because when you multiply it you should be able to get a large number.
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3 years ago
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62.069% is what i thank it is
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The equation for a proportional relationship is y =5.8x. The graph of the relationship passes through the point (1,5.8) which re
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Answer:

i neeeddddd poitndaugsins

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