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aivan3 [116]
2 years ago
15

The equipment account had a $36,000 balance at the beginning of the year, and a $30,000 balance at the end of the year. The accu

mulated depreciation account had a balance of $22,000 at the beginning of the year, and a $17,000 balance at the end of the year. The income statement reported depreciation expense of $4,000 for the year. Equipment costing $10,000 was sold for its book value. Cash received from the sale to be reported in the Investing Activities section is $
Business
1 answer:
MrMuchimi2 years ago
6 0

Answer:1000

Explanation:

Equipment decreases $6000 ($10000-$4000). Accumulated depreciation decreases $9000 ($22000+4000-$17000). $10000 cost -$9000 accumulated depreciation = $1000 cash received from sale.

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Terrell has an employer-sponsored 401(k) plan that he contributes to, and his employer matches 25% of his 401(k) contributions.
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The participants in a television quiz show are picked from a large pool of applicants with approximately equal numbers of men an
IgorLugansk [536]

Answer:

a) P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)

P(X=0)=(11C0)(0.5)^0 (1-0.5)^{11-0}=0.00049

P(X=1)=(11C0)(0.5)^1 (1-0.5)^{11-1}=0.0054

P(X=2)=(11C0)(0.5)^2 (1-0.5)^{11-2}=0.027

And adding we got:

P(X \leq 2)= 0.033

b) P(X \geq 2)= 1-P(X

And replacing we got:

P(X \geq 2)= =1-[0.00049 +0.0054] = 0.994

c) P(X \leq 1)= 1-P(X

And replacing we got:

P(X \leq 1)=0.00049 +0.0054= 0.0059

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest "number of women", on this case we now that:

X \sim Binom(n=11, p=0.5)

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

For this case we want to find this probability:

P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)

P(X=0)=(11C0)(0.5)^0 (1-0.5)^{11-0}=0.00049

P(X=1)=(11C0)(0.5)^1 (1-0.5)^{11-1}=0.0054

P(X=2)=(11C0)(0.5)^2 (1-0.5)^{11-2}=0.027

And adding we got:

P(X \leq 2)= 0.033

Part b

For this case we want this probability:

P(X \geq 2)

And we can use the complement rule and we got:

P(X \geq 2)= 1-P(X

And replacing we got:

P(X \geq 2)= =1-[0.00049 +0.0054] = 0.994

Part c

For this case we want this probability:

P(X \leq 1)

And we can use the complement rule and we got:

P(X \leq 1)= 1-P(X

And replacing we got:

P(X \leq 1)=0.00049 +0.0054= 0.0059

6 0
1 year ago
Suppose that a chicken farm uses a nearby stream to dispose of the wastes released by its chickens. These wastes flow downstream
Bess [88]

Answer:

See attached photo.

Explanation:

Refer to the photo attached.

7 0
2 years ago
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