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marshall27 [118]
3 years ago
6

What is 15-10c=13 worked out

Mathematics
1 answer:
exis [7]3 years ago
7 0

Answer:

15 - 10c = 13 \\ 10c = 15 - 13 \\ 10c = 2 \\ c =  \frac{2}{10}  \\ c =  \frac{1}{5}

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The equation of the area of a triangle is Area=Length×Width. We know area=256.5 and length is 18. So now we know 256.5=18×width. We then divide 256.5÷18 to get the width. The width is 14.25. Perimeter=legth+length+width+width. If width is 18 and length is 14.25 then we add up all 4 sides. 18+18+14.25+14.25=64.5. 64.5 is the perimeter
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Solve photo question. Mathematics. Please. Easy one. And last one.
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A collection of nickels and dimes is worth $4.40. There are 53 coins in all. How many nickels are there?

Nickel is a US coin worth 5 cents or 0.05.
Dime is a US coin worth 10 cents or 0.10

n + d = 53
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n = 53 - d
0.05(53 - d) + 0.10d = 4.40  
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4 0
3 years ago
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I need help with this question.
aliina [53]

Answer:

answer below

Step-by-step explanation:

in this case the decimal will be recurring so 1 and 2 thirds is 1.6 recurring and 2 and 7 nigths is 2.7 recurring so just plot in between the intervals so the first one could be between 1.6 and 1.7 and the sexond between 2.7 and 2.8

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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:

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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.

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Answer: 2500

Step-by-step explanation:

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