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4 years ago

A 9kg bag of mangoes for $12 is

1 answer:

3 0

A 9 kg bag of mangoes for $12 is. Given is the bulk unit rate of the mangoes. Let’s solve for the unit rate of each mango and its price. => Let’s analyze the given situation. It is said that in 9 kilograms of bag of mangoes, the costs of all of it is 12 dollars. Find how much the cost of 1 kg bag of mangoes is. => 12 dollars divided by 9 kilograms to get the value of 1kg bag of mangoes => 12 / 9 = 1.3 Thus, the unit rate is 1.3 dollars / kg

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How would I give 12.50 dollars to 350 people evenly

Nadya [2.5K]

Answer:

12.5/350

Step-by-step explanation:

12.50 amount of money divided by amount of people

12.5/350

roughly .035 per person but half cents don't exist

3 0

3 years ago

I need help! Is this correct

riadik2000 [5.3K]

Answer:

Yes

Step-by-step explanation:

5 0

3 years ago

Find the total composite area

prohojiy [21]

Answer:

Times, and conquer.

Step-by-step explanation:

First half the picture, for the circle and square. so we know the base is 28m, and the length including the circle is 24. So now times the two numbers and bam your answer.

3 0

3 years ago

Evaluate 10 m + n 2 4 10m+ 4 n 2 10, m, plus, start fraction, n, squared, divided by, 4, end fraction when m = 5 m=5m, equals,

Bond [772]

Answer:

$\therefore10m+\frac{n^2}4 =54$

Step-by-step explanation:

Variables : The value of a quantity which can be changed.

Monomial : Monomial contains only one term.

Example 3x, 4,6y² etc.

Binomial: Binomial contains two terms.

Example 3+4y, 20+7z etc

Trinomial: Trinomial contains 3 terms.

Example 5x-2y+5, 7z-3y+4x etc.

Given that,

$10m+\frac{n^2}4$ [ It is a binomial.]

Putting m =5 and n=4

$= (10\times 5)+\frac{4^2}4$

$=50+\frac{4\times4}{4}$

$=50+4$

=54

$\therefore10m+\frac{n^2}4 =54$

6 0

4 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.

4 0

4 years ago