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3 years ago

6

If there is 2 bands and one as 56 members and the other has 96 members and needs to be lined up in equal rows how much members w

ill be in each row

1 answer:

dlinn [17]3 years ago

7 0

if there are 2 rows then each row will have 78 members

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you have a bag with 20 marbles 2 yellow 5 green 12 red and 1 blue which of the following statements are true? Select all that ap

NNADVOKAT [17]

An event has probability 1 if you're certain that it will happen. You are sure to pick a marble that's either yellow, green, red, or blue. So, this probability is 1.
There are more yellow marbles than blue marbles, so it's actually more likely that you'll pick a yellow one, rather than the blue one.
The probability of picking a green marble is given by the number of green marbles divided by the total number of marbles. Compute this ratio and see what you get.
Similarly, the probability of picking a black marble is given by the number of black marbles divided by the total number of marbles. There are no black marbles, so...

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3 years ago

The scores on the verbal section of the Graduate Records Examination (GRE) are approximately normally distributed with a mean of

Mashcka [7]

Answer:

0.039

Step-by-step explanation:

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3 years ago

Find upper and lower limit of a weight of a cotton reel which is 1.56g nearst to mg

Papessa [141]

Answer: the lower limit is 1.555g and the upper limit is 1.564g

Step-by-step explanation:

a mg is:

1mg = 0.001g

Then we need to look at the third digit after the decimal point.

Then the weight 1.56g is rounded around this.

Remember that if the third digit after the decimal point is 5 or bigger, then we round up.

if the third digit is smaller than 5, then we round down.

Now, the maximum possible value of this weight is when the third digit is equal to 4 (4 mg) where because it is smaller than 5, we round it down to 1.56g

Then the maximum is 1.564g

And the minimum value is when we have:

w = 1.555g

Because the third digit is a 5, we round it up to 1.56g.

So the lower limit is 1.555g and the upper limit is 1.564g

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3 years ago

Jacobs & Johnson, an accounting firm, employs 19 accountants, of whom 7 are CPAs. If a delegation of 2 accountants is random

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Answer:

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Step-by-step explanation:

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3 years ago

Suppose that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to

attashe74 [19]

Answer:

0.9726 = 97.26% approximate probability that X is at most 30

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

11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped).

This means that $p = 0.11$

Random sample of 200 shafts

This means that $n = 200$

Mean and Standard deviation:

$\mu = E(x) = np = 200*0.11 = 22$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.11*0.89} = 4.42$

(a) What is the (approximate) probability that X is at most 30

Using continuity correction, this is $P(X \leq 30 + 0.5) = P(X \leq 30.5)$ , which is the pvalue of Z when X = 30.5. So

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

$Z = \frac{30.5 - 22}{4.42}$

$Z = 1.92$

$Z = 1.92$ has a pvalue of 0.9726.

0.9726 = 97.26% approximate probability that X is at most 30

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3 years ago