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siniylev [52]
2 years ago
9

47 lbs = ______________ kgs (give answer in decimal form)

Mathematics
2 answers:
creativ13 [48]2 years ago
8 0
1 pound is equal to about 0.45 kilograms.  47 lbs would be about 21.31 kilograms.
Lyrx [107]2 years ago
4 0

"lbs" (pounds) is a measure of force, whereas "kgs" (kilograms) is
a measure of mass.  Being different units with different physical
dimensions, they can never be 'equal'.

-- 47 lbs is the weight of 64.9 kgs of mass on Mars.
-- 47 lbs is the weight of 128.8 kgs of mass on the Moon.
-- 47 lbs is the weight of 21.3 kgs of mass on Earth.
.
.
etc.

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Which of the following is NOT true when testing a claim about a​ proportion? Choose the correct answer below. A. Both the tradit
morpeh [17]

Answer: C. A conclusion based on a confidence interval estimate will be the same as a conclusion based on a hypothesis test.

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To write a null hypothesis, first, start by asking a question. Rephrase that question in a form that assumes no relationship between the variables. In other words, assume a treatment has no effect. Write your hypothesis in a way that reflects this.

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5 0
2 years ago
A machine can manufacture 24,000 plastic balls in 8 hours. Find the unit rate in balls per hour.
Lorico [155]
24,000/8= 3,000 balls

The machine produce 3,000 balls per hour.

Hope that helps!
5 0
2 years ago
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Let Y denote a geometric random variable with probability of success p. a Show that for a positive integer a, P(Y > a) = qa .
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Answer:

a) For this case we can find the cumulative distribution function first:

F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k

So then by the complement rule we have this:

P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a

b) P(Y>a)= q^a

P(Y>b) = q^b

So then we have this using independence:

P(Y> a+b) = q^{a+b}

We want to find the following probability:

P(Y> a+b |Y>a)

Using the definition of conditional probability we got:

P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)

And we see that if a = 2 and b=5 we have:

P(Y> 2+5 | Y>2) = P(Y>5)

c) For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

P(X=x)=(1-p)^{x-1} p

If we define the random of variable Y we know that:

Y\sim Geo (1-p)

Part a

For this case we can find the cumulative distribution function first:

F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k

So then by the complement rule we have this:

P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a

Part b

For this case we can use the result from part a to conclude that:

P(Y>a)= q^a

P(Y>b) = q^b

So then we have this assuming independence:

P(Y> a+b) = q^{a+b}

We want to find the following probability:

P(Y> a+b |Y>a)

Using the definition of conditional probability we got:

P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)

And we see that if a = 2 and b=5 we have:

P(Y> 2+5 | Y>2) = P(Y>5)

Part c

For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

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MA_775_DIABLO [31]
The answer is c because is a vrtical line



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2 years ago
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Torrey starts a new job with an annual salary of $60,000. For each year she continues to work for the same company, she will rec
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The third choice will work
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1 year ago
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