I forgot how to do long division

They are independent because the cards are always going to stay the same. If each card that was picked and discarded, that would alter the future which would make it dependent. But... since it doesn't change the cards, it is independent.

Answer: independent

5 0

3 years ago

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Margaret has 16 yards of fabric and she is making dresses for z doll write the expression for the length of fabric used for each

emmasim [6.3K]

<h3>The expression for the length of fabric used for each doll is $\frac{16}{z}\ yards$ </h3>

<em><u>Solution:</u></em>

Given that,

Margaret has 16 yards of fabric

she is making dresses for z doll

Therefore,

Total length of fabric = 16 yards

Number of dresses = z dolls

Therefore,

$\text{ length of fabric used for each doll } = \frac{ \text{ Total length of fabric }} {\text{ z dolls }}\\\\\text{ length of fabric used for each doll } = \frac{16}{z}\ yards$

Thus, the expression for the length of fabric used for each doll is $\frac{16}{z}\ yards$

7 0

3 years ago

Consider the probability that greater than 96 out of 153 DVDs will work correctly. Assume the probability that a given DVD will

ICE Princess25 [194]

Answer:

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .