Would the answer be 12.5

13) To finanace her first car, Maria takes out a

Elodia [21]

$300

3%=.03 of something
Find .03 of 10000
10000x.03=$300

5 0

3 years ago

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What is (-6) to the negative 5 exponent over 3 to the negative 2 exponent?

loris [4]

Answer:

$\displaystyle -\frac{1}{864}$

Step-by-step explanation:

$\displaystyle \frac{-6^{-5}}{3^{-2}} = \frac{-\frac{1}{7776}}{\frac{1}{9}} = -\frac{1}{864}$

8 0

3 years ago

What is this answer I dont know most of it

RideAnS [48]

You just divide 79 into 4,032. Have you learned long division yet?

5 0

3 years ago

Simplify the expression 7x+12-4x-3

Leya [2.2K]

Answer:

3x+9

Step-by-step explanation:

Hope this helped!

5 0

3 years ago

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Two types of coins are produced at a factory: a fair coin and a biased one that comes up heads 60 percent of the time. We have o

liraira [26]

Answer:

i) 0.1% probability that if the coin is actually fair, we reach a false conclusion.

ii) 0.05% probability that if the coin is actually unfair, we reach a false conclusion

Step-by-step explanation:

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:

Fair coin:

Comes up heads 50% of the time, so $p = 0.5$

1000 trials, so $n = 1000$

So

$E(X) = np = 1000*0.5 = 500$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.5*0.5} = 15.81$

If the coin lands on heads 550 or more times, then we shall conclude that it is a biased coin.

(i) If the coin is actually fair, what is the probability that we shall reach a false conclusion?

This is the probability that the number of heads is 550 or more, so this is 1 subtracted by the pvalue of Z when X = 549.

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

$Z = \frac{549 - 500}{15.81}$

$Z = 3.1$

$Z = 3.1$ has a pvalue of 0.9990

1 - 0.9990 = 0.001

0.1% probability that if the coin is actually fair, we reach a false conclusion.

(ii) If the coin is actually unfair, what is the probability that we shall reach a false conclusion?

Comes up heads 60% of the time, so $p = 0.6$

1000 trials, so $n = 1000$

So

$E(X) = np = 1000*0.6 = 600$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.6*0.4} = 15.49$

If the coin lands on less than 550 times(that is, 549 or less), then we shall conclude that it is a biased coin.

So this is the pvalue of Z when X = 549.

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

$Z = \frac{549 - 600}{15.49}$

$Z = -3.29$

$Z = -3.29$ has a pvalue of 0.0005

0.05% probability that if the coin is actually unfair, we reach a false conclusion

5 0

3 years ago