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3 years ago

Number 5. I neeed the answer please and thankyou

Mathematics

2 answers:

adell [148]3 years ago

6 0

Answer:

0.4 or 2/5's of a loaf of bread

Step-by-step explanation:

For example, in number 3 the number is higher than 1 because there's more burritos than friends to share them with. In #5. the question tells you that there's 2 loaves of bread but 5 bakers meaning that each baker will have less than 1 loaf of bread. 2 divided by 5 is 2/5 or 0.4

liberstina [14]3 years ago

4 0

2/5.
2 loaves of bread is divided amongst 5 people. so 1 person gets less than a loaf of bread.

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Complementary & Su Find the value of x. 1) (4x + 3) 77° X=

ddd [48]

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x=20

Step-by-step explanation:

x=20 because (4x+3)=77 so you take out of parnetheses

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2 years ago

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A very large batch of parts (assume a normal distrbution0 from a manufacturer has a mean weight = 43g and a standard deviation =

AVprozaik [17]

Answer:

5.44% probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Percentage of parts with weights exceeding 45g?

1 subtracted by the pvalue of Z when X = 45. So

We have $\mu = 43, \sigma = 4$

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

$Z = \frac{45 - 43}{4}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

1 - 0.6915 = 0.3075

If you slect 16 parts at random form that batch, what is the probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g?

This is P(X = 8) when n = 16, p = 0.3075. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{16,8}.(0.3075)^{8}.(0.6915)^{8} = 0.0544$