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dedylja [7]
2 years ago
8

one pound of swordfish costs as much as 1.5 pounds of salmon. Mrs. O pays $39 for 2 pounds of salmon and 3 pounds of swordfish.

Find the ratio of the amount of money Mrs. O pays for the swordfish to the amount of money she pays for the swordfish.
Mathematics
2 answers:
myrzilka [38]2 years ago
7 0
<h2>Answer with explanation:</h2>

It is given that:

one pound of swordfish costs as much as 1.5 pounds of salmon.

Let x be the cost of one pound of salmon.

This means that the cost of one pound of swordfish is: 1.5x

Mrs. O pays $39 for 2 pounds of salmon and 3 pounds of swordfish.

This means that:

2x+3(1.5x)=39\\\\2x+4.5x=39\\\\6.5x=39\\\\x=\dfrac{39}{6.5}\\\\x=6

This means that cost of 1 pound of salmon= $ 6

Hence, cost of 2 pounds of salmon= $ 12  ( since, 6×2=12 )

and cost of 1 pound of swordfish= $ (1.5×6)= $ 9

Hence,

cost of 3 pound of swordfish= $ (9×3)=$ 27

  • Hence, ratio of amount of money she pays for salmon to the swordfish is:

                    12/27= 4/9

                         i.e.   4:9

  • and the ratio of  amount of money she pays for swordfish to the salmon is:

                         9:4

Lostsunrise [7]2 years ago
4 0
For every 2 pounds of salmon she pays for she will get 3 pounds of sword fish with it. 2 to 3 or 2/3 are different ways to Write a ratio for this problem.
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If x and y satisfy both 9x+2y=8 and 7x+2y=4, then y=?
Alex

Answer:

The value of y is -5

Step-by-step explanation:

we have

9x+2y=8 ------> equation A

7x+2y=4 ------> equation B

we know that

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Using a graphing tool

Remember that

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therefore

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2 years ago
The distribution of the number of moths captured per night by a certain moth trap is approximately normal with mean 103. If 28 p
Afina-wow [57]

Answer:

0.28 = (76 - 103)/σ (B)

Step-by-step explanation:

Let X be a random variable for capturing moths every night

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σ is the standard deviation

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Pr(X<76) = 0.28

For normal distribution, Z = (X - μ)/σ

Pr(X<76) = (X - μ)/σ

0.28 = (76 - 103)/σ

The equation to find the standard deviation is

0.28 = (76 - 103)/σ

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Answer:

P(z>-1.768)=1-P(z

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the scores for the univerisity A, and we know that:

X \sim N (\mu = 625,\sigma =\sqrt{100}= 10)

Let Y the scores for the univerisity B, and we know that:

Y \sim N (\mu = 600,\sigma =\sqrt{150}= 12.25)

We select a sample size of size n=2, and since the distirbution for X is normal then the distribution for the sample mean would be given by:

\bar X \sim N(\mu=625, \frac{\sigma}{\sqrt{n}}=\frac{10}{\sqrt{2}}=7.07)

And for the univeristy B we select a sample of n=3

\bar Y \sim N(\mu=600, \frac{\sigma}{\sqrt{n}}=\frac{12.25}{\sqrt{3}}=7.07)

Since both sample means are normally distributed then the difference Z= \bar X- \bar Y is also normal distributed with the following parameters:

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Z= \frac{z -\mu_z}{\sigma_z}

And if we replace we got :

Z= \frac{0-25}{14.14}=-1.768

And if we find the probability using the normla standard table or excel we got:

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3 0
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