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Darya [45]
3 years ago
11

SOMEONE HELP ME I NEED ANSWERS!!!!!!!!!!!!!!!

Mathematics
2 answers:
zaharov [31]3 years ago
8 0
Okay calm down i have the answer, it is point d <span />
cupoosta [38]3 years ago
8 0
C) between D and A turtle power
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What is the factor x²+6x+​
Lera25 [3.4K]

Answer:

this is what i found

Factor  

x  out of  x  2 .  x ⋅  x −  6 x  Factor  x  out of  − 6 x .  x ⋅ x  + x ⋅ − 6  Factor  x  out of  x ⋅ x + x ⋅ − 6 . x ( x − 6  )

8 0
2 years ago
You buy a car for $16,000. It will decrease in value by 10% each year. How much will it be worth after 4 years?
Mandarinka [93]

Answer: 9,600

Step-by-step explanation:

I could possibly did wrong but I shall explain this none the less. I first knew that 10% of 16,000 was 1,600. So then I just timed that by four getting 6,400. I then did 16,000-6,400 getting $9,600. Sorry if I’m wrong

7 0
3 years ago
Read 2 more answers
Factor the expression. 35g2 – 2gh – 24h2
klasskru [66]
A. 7g-5g=6-4
2g=2
g=2/2
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7 0
3 years ago
Read 2 more answers
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
Help me please , I need to turn this in by midnight
meriva

Answer:

Step-by-step explanation:

for IV you do 40^2+30^2=c, or 1600+ 900=c. c= 2500 and unsquare that you get 50

8 0
3 years ago
Read 2 more answers
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