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2 years ago

What is the missing numerator?

Mathematics

1 answer:

N76 [4]2 years ago

3 0

Answer:

Step-by-step explanation:

As with any other addition fact problem, you can isolate the unknown term by subtracting the unwanted constant.

5/7 +x/6 = 51/42

x/6 = 51/42 -5/7 . . . . . subtract 5/7 from both sides

x/6 = 51/42 -30/42 = 21/42 = 1/2 . . . . . complete the subtraction

x = 6(1/2) = 3 . . . . . . . . multiply by 6

The missing numerator is 3.

_____

<em>Alternate solution</em>

Multiply the equation by 42:

30 +7x = 51

7x = 51 -30 = 21 . . . . . subtract 30

x = 21/7 = 3 . . . . . . . . divide by the coefficient of x

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2 years ago

A researcher took a poll of people in his state. He asked each person if he or she had any pets (Yes or No) and to state his or

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Answer:

DF = 2

Step-by-step explanation:

Formula for degree of freedom is;

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3 years ago

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.

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3 years ago

In triangle abc, m of acb = 90, cd is perpendicular to ab , m of acd is 60. and bd is 5 cm. find ad

weeeeeb [17]

Let us draw a picture to make the things more clear.

Attached is the image.

We have been given that

$\angle acd = 60 ^{\circ}$