Jacob used 1/7 of a liter of water to fill 1/9 of the fish aquarium. How many liters r needed to fill the aquarium?

Answer:

Options 1,2 and 4 are correct.

Step-by-step explanation:

We have been given a parallelogram MNPQ was dilated to create parallelogram  M'N'P'Q' and we asked to find out which of the given options are correct.

Let us see all the options one by one.

On using distance formula we can see that side MN is 2 units.

[tex]MN=\sqrt{(x_2-x_1)^{2} +(y_2-y_1)^{2} }[/tex]

[tex]MN=\sqrt{(4-2)^{2} +(-2-(-2))^{2} }[/tex]

[tex]MN=\sqrt{(2)^{2} } =2[/tex]

Therefore,option 1 is correct.

We can find the side M'N' by distance formula as we solved for side MN. After solving for M'N' we will get 5 units. Therefore, option 2 is also correct.

Now let us see third option. We can see that image M'N'P'Q' is bigger than pre-image MNPQ. Therefore, option 3 is false.

Now look at our option 4.

Since we know slope is rise (difference of y coordinates) over run (difference of x coordinates) . Upon looking at our image and pre-image we can see that as we increase our x coordinate by one unit our y coordinate also increase by one.

Therefore, option 4 is correct indeed.

Upon looking at our fifth option we can see that our image is bigger than our pre-image. If our image was dilated by factor of 2/5 it would have been smaller than our pre-image. We can see the scale factor is 5/2 instead of 2/5.

Therefore, our fifth option is false.

Considering the slopes and scale factor of the diagram given, the statements that are true about the parallelogram are: A, B, and D.

Slope = vertical distance / horizontal distance = change in y / change in x.

Scale factor = length of image/corresponding length of pre-image.

Find the length of MN:

MN = |4 - 2| = 2 units (Option A is true).

Find M'N':

M'N' = |10 - 5| = 5 units (Option B is true).

MNPQ is the pre-image, while M'N'P'Q' is the image, therefore, the image is bigger than the pre-image. (Option C is NOT true).

Find Slope of MQ:

Slope = 1/1 = 1

Find Slope of M'Q':

Slope = 2/2 = 1 (Option D is true).

Find the scale factor:

Scale factor = dimension of image/dimension of pre-image = M'N'/MN

Scale factor = 5/2 (Option E is NOT true).

Therefore, considering the slopes and scale factor of the diagram given, the statements that are true about the parallelogram are: A, B, and D.

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Final answer:

The answer explains how to create a Python application to simulate rolling two dice using the random module.

Explanation:

application

To simulate the rolling of two dice using an object of class random, you can create a Python application. Here's an example of how you can achieve this:

Import the random module.

Create a function to simulate rolling a die.

Call the function twice to simulate rolling two dice.

Jed and Alex do not shoot free throws at the same rate, Jed is faster by 0.5 seconds.

A unitary method is a mathematical way of obtaining the value of a single unit and then deriving any no. of given units by multiplying it with the single unit.

Given, Jed took 2 free throws in 5 seconds.

∴ Jed can do 1 free throw in (5/2) = 2.5 seconds.

Alex took 4 free throws in 12 seconds.

∴ Alex can do 1 free throw in (12/4) = 3 seconds.

No, Jade and Alex do not shoot free throws at the same rate Jed is faster by 0.5 seconds.

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Answer:

A

Step-by-step explanation:

y = something x plus or minus a number

going through each line it isn't a linear pattern ( they don't all have the same difference when subtracted or added together)

so now you play around with quadratic equations ( square roots, powers, etc.)

 and you will see that y = x^2 -5

The bases are not the same, so you cannot set 3x - 1 equal to 8.

You can evaluate the logarithm on the right side of the equation to get 3.

You can use the definition of a logarithm to write 3x - 1 = 1000.

Answer:  The solution of the given equation is [tex]x=333\dfrac{2}{3}[/tex] and the value of (3x - 1) is 1000.

Step-by-step explanation:  We are given to describe the steps in solving the following logarithmic equation:

[tex]\log(3x-1)=\log_28.[/tex]

Also, we are to find the value of [tex](3x-1).[/tex]

We will be using the following logarithmic properties:

[tex](i)~\log_ba=x~~~\Rightarrow a=b^x,\\\\(ii)~\log_ba=\dfrac{\log a}{\log b},\\\\(iii)~\log a^b=b\log a.[/tex]

We note here that if the base of logarithm is not mentioned, then we assume it to be 10.

The solution is as follows:

[tex]\log(3x-1)=\log_28\\\\\Rightarrow \log(3x-1)=\dfrac{\log8}{\log2}\\\\\Rightarrow log(3x-1)=\dfrac{\log2^3}{\log2}\\\\\Rightarrow \log(3x-1)=\dfrac{3\log2}{\log2}\\\\\Rightarrow \log(3x-1)=3\\\\\Rightarrow 3x-1=10^3\\\\\Rightarrow 3x-1=1000\\\\\Rightarrow 3x=1001\\\\\Rightarrow x=\dfrac{1001}{3}\\\\\Rightarrow x=333\dfrac{2}{3}.[/tex]

Thus, the solution of the given equation is [tex]x=333\dfrac{2}{3}[/tex] and the value of (3x - 1) is 1000.

The probability that all five randomly selected sea anemones from a population with normally distributed basal diameters have a diameter more than 5.5 cm is approximately 4.5%.

To solve this problem, we first need to find the probability that one sea anemone has a basal diameter more than 5.5 cm, given the population's mean basal diameter is 5.3 cm with a standard deviation of 1.8 cm. We calculate the Z-score for a diameter of 5.5 cm to see how many standard deviations this value is away from the mean. The Z-score formula is Z = (X - μ) / σ, where X is the value in question, μ is the mean, and σ is the standard deviation.

The Z-score for a diameter of 5.5 cm is Z = (5.5 - 5.3) / 1.8 ≈ 0.11. Looking this up in a standard normal distribution table, we find that the probability of selecting an anemone with a diameter larger than 5.5 cm is approximately 0.5438. Since the selections are independent, the probability of all five having a diameter more than 5.5 cm is (0.5438)⁵ ≈ 0.045, or 4.5%.

The probability that all five randomly selected anemones have a basal diameter more than 5.5 cm is approximately 0.0056.

To solve this problem, we can use the standard normal distribution. First, we need to standardize the diameter value using the z-score formula:

[tex]z = \frac{x - \mu}{\sigma}[/tex]

where:

- x is the value of the basal diameter,

- [tex]\mu[/tex] is the mean of the basal diameters (5.3 cm),

- [tex]\sigma[/tex] is the standard deviation of the basal diameters (1.8 cm), and

- z is the standardized score.

For a basal diameter of 5.5 cm, the z-score would be:

[tex]z = \frac{5.5 - 5.3}{1.8} = \frac{0.2}{1.8} \approx 0.1111[/tex]

To find the probability that an anemone has a basal diameter more than 5.5 cm, we need to find the area to the right of z = 0.1111 under the standard normal curve.

Using a standard normal table or calculator, we find that this area is approximately 0.4562.

Since each selection is independent (assuming replacement), the probability that all five anemones have a basal diameter more than 5.5 cm is the probability raised to the power of 5:

[tex]P(\text{all five} > \text{ 5.5 cm}) = (0.4562)^5 \approx 0.0056[/tex]

So, the probability that all five randomly selected anemones have a basal diameter more than 5.5 cm is approximately 0.0056.

The correct transformation is to reflect the graph about the y-axis and then translate it 2 units down, corresponding to option 4.

The correct transformation is therefore to reflect the graph about the y-axis to change the slope from positive to negative and then translate 2 units down to change the y-intercept from +9 to +7. This corresponds to option 4 from the question.

The requried fraction of the bouquet that is pink roses is 7/12, as of the given proportion.

If Angelica's bouquet of a dozen roses contains 5 white roses, then the remaining roses must be pink. Since there are 12 roses in a dozen, the number of pink roses in the bouquet would be 12 - 5 = 7.

To find the fraction of the bouquet that is pink roses, we divide the number of pink roses by the total number of roses:

Fraction of pink roses = Number of pink roses / Total number of roses

Fraction of pink roses = 7 / 12

Therefore, the fraction of the bouquet that is pink roses is 7/12.

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The expression 7 x 10³ is 3.5 times larger than 2 x 10³ because the power of 10 is the same in both, so we simply compare the coefficients 7 and 2.

The student is asking to compare two expressions written in scientific notation: 7 x 10³ and 2 x 10³. Scientific notation allows us to write very large or very small numbers conveniently and is commonly taught in middle school math.

To compare these two expressions, we can look at the coefficients (7 and 2) because the powers of 10 are the same in both expressions (10³). Since these powers are the same, comparing the coefficients directly will give us the relationship between the two expressions. Here, 7 is simply 3.5 times as large as 2, so we can say that 7 x 10³ is 3.5 times larger than 2 x 10³.

In terms of the answer choices provided by the student, we can thus conclude that the correct answer is:

The line will shift vertically down by 4 tons.

The simplest method to do this is with the vector dot product. Let the vector A = <9i + 5j> with magnitude √106 be the base diagonal, and B = <5j + 3k> be the diagonal vector on the side, with magnitude √34. Then cos θ = (A dot B) divided by the product of the magnitudes. A dot B =30, so 

cos θ = 30 / √(34 x 106)

 = 0.4997 ==>

 θ = 49.19° is the answer