Can someone please help mee :)

Mathematics

1 answer:

Lunna [17]3 years ago

5 0

Simplify the complex fractions
A) (from the picture) = 1 2/5.
B) (from the picture) = 13/22
Solve each equation below
A) (from the picture) x= 9
B) (from the picture) w= 10 1/2
C) (from the picture) y= -80
Find the sum of each number below. Describe how you know what the sign of you answer will be.
A) -19 + 8 = -11
B) -6 + (-5) = -11

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Triangle ABC has the following coordinates: A(4 , 5), B(5 , 3), and C(2 , 3). Part A: If triangle ABC is dilated by a scale fact

Kobotan [32]

The new coordinates of A'B'C' creates a triangle that is larger than ABC.

<h3>Transformation</h3>

Transformation is the movement of a point from its initial location to a new location. Types of transformation are <em>translation, reflection, rotation and dilation.</em>

If a point A(x, y) is dilated by a scale factor k, the new point is at A'(kx, ky).

Given that:

Triangle ABC has the following coordinates: A(4 , 5), B(5 , 3), and C(2 , 3)

If it is dilated by a scale factor of 3, the new point is at:

A'(12, 15), B'(15, 9) and C'(6, 9)

Therefore the new coordinates of A'B'C' creates a triangle that is larger than ABC.

Find out more on dilation at: brainly.com/question/10253650

8 0

3 years ago

At Pike Place Fish Market in Seattle, customers can purchase a variety of different types of seafood. One type of seafood sold a

allsm [11]

Answer:

$z=1.28$

And if we solve for a we got

$a=3.6 +1.28*0.8=4.624$

So the value of height that separates the bottom 90% of data from the top 10% is 4.624.

So then the best answer for this case would be:

C. 4.64

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 random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(3.6,0.8)$

Where $\mu=3.6$ and $\sigma=0.8$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$