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2 years ago

SOMEEEEONEEE pls help pls and thx :)

1 answer:

7 0

1) The sculptor created a marble basin with an approximate volume of 94,509.6 cubic centimeters - Option D, 2) The approximate area of the heart-shaped cake is 283 square inches - Option B, 3) The length of the wooden frame around the window is 94.3 inches - Option B.

In this question we should make use of geometric formulas for volumes and areas and key information from statement in order to find the right choices.

1) In this case, the volume occupied by the marble water basin ( $V$ ), in cubic centimeters, by subtracting the volume of the hemisphere from the volume of the cylinder:

$V = \pi\cdot R^{2}\cdot h - \frac{2\pi}{3} \cdot r^{3}$ (1)

Where:

$R$ - Radius of the cylinder, in centimeters.
$h$ - Height of the cylinder, in centimeters.
$r$ - Radius of the cylinder, in centimeters.

If we know that $R = 30\,cm$ , $h = 45\,cm$ and $r = 25\,cm$ , then the volume of the marble is:

$V = \pi \cdot (30\,cm)^{2}\cdot (45\,cm) - \frac{2\pi}{3}\cdot (25\,cm)^{3}$

$V \approx 94509.579\,cm^{3}$

The right choice is D.

2) To determine the approximate surface area of the cake covered in red frosting ( $A_{s}$ ), in square inches, we need to find the sum of the surface area of the entire circle and the surface area of the square:

$A_{s} = l^{2} + 2\cdot l\cdot h + \frac{\pi}{4}\cdot D^2 + \pi \cdot D\cdot h$ (2)

Where:

$l$ - Side length of the square, in inches.
$h$ - Height of the cakes, in inches.
$D$ - Diameter of the cakes, in inches.

If we know that $l = 9\,in$ , $h = 3\,in$ and $D = 9\,in$ , then the <em>approximate</em> area covered in red frosting:

$A_{s} = (9\,in)^{2} + 2\cdot (9\,in)\cdot (3\,in) + \frac{\pi}{4}\cdot (9\,in)^{2} + \pi \cdot (9\,in)\cdot (3\,in)$

$A_{s} \approx 283.440\,in^{2}$

The right choice is B.

3) The frame around the window is found by means of the following perimeter formula ( $s$ ), in inches:

$s = \pi\cdot r + 2\cdot h + 2\cdot r$ (3)

Where:

$r$ - Radius, in inches.
$h$ - Height of the rectangle, in inches.

If we know that $r = 9\,in$ and $h = 24\,in$ , then the length of the frame around the window is:

$s = \pi\cdot (9\,in) + 2\cdot (24\,in) + 2\cdot (9\,in)$

$s \approx 94.274\,in$

The right choice is B.

We kindly invite to see question on volumes: brainly.com/question/1578538

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Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribu

Svetllana [295]

Answer:

Step-by-step explanation:

Hello!

You have the variable

X: Area that can be painted with a can of spray paint (feet²)

The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²

since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.

P(X>27)

First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of $P(Z$ , you have to do the following conervertion:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².

The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:

X[bar]~N(μ;δ²/n)

So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)

To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.

I hope it helps!

4 0

2 years ago

During batting practice, two pop flies are hit from the same location, 2 s apart. The paths are modeled by the equations h = -16

jolli1 [7]

To find the time at which both balls are at the same height, set the equations equal to each other then solve for t.
h = -16t^2 + 56t
h = -16t^2 + 156t - 248
-16t^2 + 56t = -16t^2 + 156t - 248
You can cancel out the -16t^2's to get
56t = 156t - 248
=> 0 = 100t - 248
=> 248 = 100t
=> 2.48 = t
Using this time value, plug into either equation to find the height.
h = 16(2.48)^2 + 56(2.48)
Final answer:
h = 40.4736
Hope I helped :)

6 0

2 years ago

Read 2 more answers

Randy presses RAND on his calculator twice to obtain two random numbers between 0 and 1. Let $p$ be the probability that these t

anygoal [31]

Answer:

$\frac{\pi}{4}$

Step-by-step explanation:

Lets call x,y the numbers we obtain from the calculator. x and y are independent random variables of uniform[0,1] distribution.

Lets note that, since both x and y are between 0 and 1, then 1 is the biggest side of the triangle.

Lets first make a geometric interpretation. If the triangle were to be rectangle, then the side of lenght 1 should be its hypotenuse, and therefore x and y should satisfy this property:

x²+y² = 1

Remember that in this case we are supposing the triangle to be rectangle. But the exercise asks us to obtain an obtuse triangle. For that we will need to increase the angle obtained by the sides of lenght x and y. We can do that by 'expanding' the triangle, but if we do that preserving the values of x and y, then the side of lenght 1 should increase its lenght, which we dont want to. Thus, if we expand the triangle then we should also reduce the value of x and/or y so that the side of lenght 1 could preserve its lenght. With this intuition we could deduce that

x²+y² < 1

Now lets do a more mathematical approach. According to the Cosine theorem, a triangle of three sides of lenght a,b,c satisfies

a² = b²+c² - 2bc* cos(α), where α is the angle between the sides of lenght b and c.

Aplying this formula to our triangle, we have that

$1^2 = x^2 + y^2 - 2bc* cos(\alpha)$ , where $\alpha$ is the angle between the sides of lenght x and y.

Since the triangle is obtuse, then $\pi/2 < \alpha < \pi$ , and for those values $cos(\alpha)$ is negative , hence we also obtain

1 > x² + y²

Thus, we need to calculate P(x²+y² <1). This probability can be calculated throught integration. We need to use polar coordinates.

(x, y) = (r*cosФ,r*senФ)

Where r is between 0 and 1, and Ф is between 0 and $\pi /2$ (that way, the numbers are positive).

The jacobian matrix has determinant r, therefore,

${\int\int}\limits_{x^2+y^2 < 1} \, dxdy = \int\limits^1_0\int\limits^{\frac{\pi}{2}}_0 {r} \, d\phi dr = \frac{\pi}{2} * \int\limits^1_0 {r} \, dr = \frac{\pi}{2} * (\frac{r^2}{2} |^1_0) = \frac{\pi}{4}$

As a conclusion, the probability of obtaining an obtuse triangle is $\frac{\pi}{4}$ .

6 0

2 years ago

Read 2 more answers

Jeremy had 3/4 of a submarine sandwich and gave

NeX [460]

Actual question is "<span>Jeremy had 3/4 of a submarine sandwich and gave his friend 1/3 of it. What fraction of the sandwich did the friend receive?"

Solution:
1/3 of 3/4 of a submarine sandwich = 1/3x3/4 = 3/12 = 1/4
so jermy gave 1/4 of the submarine sandwich to his friend. </span>

5 0

1 year ago

A jury pool consists of 27 people. How many different ways can 11 people be chosen to serve on a jury and one additional person

hichkok12 [17]

Answer:

208,606,320

Step-by-step explanation:

A jury pool consists of 27 people. 11 people be chosen to serve on a jury and one additional person be chosen to serve as the jury foreman

Total of 27 people. 11 people can be selected from 27 people in 27C11 ways

$27C11= \frac{27!}{11!(27-11)!} =13037895$

one additional person be chosen to serve as the jury foreman

after selecting 11 people , remaining is 27 - 11= 16 people

Now 1 is selected from 16 in 16 ways

Number of different ways = $13037895 \cdot 16=208606320$

4 0

2 years ago