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3 years ago

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There is only one city in Delaware county and that is dunston. The population of dunston is 6549. There are 43118 people living

in Delaware county but not in dunston. What is the total population of Delaware county.

1 answer:

WARRIOR [948]3 years ago

6 0

Honestly this makes no logical sense but you should take 43,118+6,549=49,667

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Vilka [71]

Answer:

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y = -3 -4 -2 -1

Step-by-step explanation:

equation : 3y+9x = -9

when x = 1

3y+3x = -9

or, 3y + 3×1= -9

or, 3y+3=-9

or, 3y = -12

so, y = -4

Again

Let's suppose x as -1

3y+3x= -9

or, 3y+ 3×-1 = -9

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or, 3y = -6

y = -2

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In basketball you score 2 points for a field goal and 1 point for a free throw. Suppose that you have scored at least 3 points i

olga55 [171]

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Solve the equation and give me how you solved it and the answer

Viktor [21]

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Tell what you would do to isolate the variable.

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3 years ago

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The Town of Hertfordshire clerk knows that 23% of dogs in the town have completed emotional support training. Hertfordshire plan

Nataly_w [17]

Answer:

95.64% probability that under 30% of the dogs are emotional support trained

Step-by-step explanation:

For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$\mu = E(X) = np = 100*0.23 = 23$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.23*0.77} = 4.21$

What is the probability that under 30% of the dogs are emotional support trained?

30% of 100 is 0.3*100 = 30

So this is the pvalue of Z when X = 30.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30 - 23}{4.1}$

$Z = 1.71$

$Z = 1.71$ has a pvalue of 0.9564.

So there is a 95.64% probability that under 30% of the dogs are emotional support trained

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