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Dmitrij [34]
3 years ago
12

Define 7,080,267 in words

Mathematics
2 answers:
REY [17]3 years ago
3 0

seven million eighty thousand two hundred sixtiy seven

Readme [11.4K]3 years ago
3 0

seven million eighty thousand two hundred sixty seven

Is that the kind of define you need?

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um am playing a game i forgot the math anwser the qusitoin says that the sum is 9 and thier diffrince is 1 what are the two numb
Virty [35]

Answer:

8

Step-by-step explanation:

All you have to do it subtract 9 from 1 than that's it your answer.

3 0
3 years ago
The populations of two cities after t years can be modeled by -150t+50,000 and 50t+70,0000 . What is the difference in the popul
Sindrei [870]

Answer:

25,800

Step-by-step explanation:

Plug in 4 everytime you see T.

multiple-150 times 4 plus 50000 which gives you 49,400.

then multiple 50 times 4 and add 75000, giving you 75,200 then subtract 75,200 - 49,400 which you end up with 25,800.

hope this helped

5 0
2 years ago
What is the measure of
Anestetic [448]
It’s either 100 or 130
8 0
3 years ago
In a business, if A can earn $7500 in 2.5 years, find the unit rate of his earning per month
andrew11 [14]

2.5 years = 30 months

7500$ in 30 months

x$ in 1 month

___________________

x = (7500*1)/30 = 250

The answer is 250$/month.

5 0
3 years ago
Read 2 more answers
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

5 0
3 years ago
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