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3 years ago

If a= b^2/3 and b= c^3/4; what is the relation between a and c?

Mathematics

1 answer:

Lorico [155]3 years ago

6 0

Answer: a = c^(1/2) which is the first answer choice of the list

Work Shown:

a = b^(2/3) ... given equation

a = (c^(3/4))^(2/3) ... replace b with c^(3/4)

a = c^((3/4)*(2/3)) .... multiply exponents

a = c^(1/2) .... Simplify; note the 3's cancel, also 2/4 turns into 1/2

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3 years ago

Sora correctly solved this inequality.

Black_prince [1.1K]

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3 years ago

Read 2 more answers

Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa

Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation

n=580 represent the random sample taken

X represent the seafood sold in the country that is mislabeled or misidentified by the people

$\hat p=0.25$ estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

$\alpha=0.01$ represent the significance level (no given, but is assumed)

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204$

$0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296$

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

Part c

A government spokesperson claimed that the sample size was too small, relative to the billions of pieces of seafood sold each year, to generalize. Is this criticism valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0

4 years ago

Can you please help me with this

Vitek1552 [10]

Answer:

log9

Step-by-step explanation:

Using the rules of logarithms

logx + logy = log(xy)

logx - logy = log( $\frac{x}{y}$ )