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Drupady [299]
3 years ago
7

What is 9(5x)? I really need help on this

Mathematics
2 answers:
zheka24 [161]3 years ago
8 0

Answer:

45x

Step-by-step explanation:

that is the answer!!! :)

icang [17]3 years ago
4 0

Answer:

45x

Explanation:

  • 9(5x)

Remove Parenthesis

9 · 5x

Multiply And Add x

45x

- PNW

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Evelyn's made a cake in the shape of a circus tent the cake consisted of a cylinder and a cone .The height of the cone is half t
Lera25 [3.4K]
The volume of the cake is 1470 in³. 
volume of a cylinder = πr² x height
(Think about how a cylinder is basically a bunch of circles stacked on top of each other. To find the volume, first you need the area of the circle (πr², then you multiply by how many circles you are stacking on top of each other (height)) 

we know the diameter of the cylinder is 12 in. and the radius is half of the diameter. 
half of 12 is 6, therefore the radius is 6 in. or r = 6 
Assuming pi is 3.14, solve for the height of the cylinder
1470 = (3.14)(6²)(height)
1470 = 3.14 x 36 x height
1470 = 113.04 x height
height ≈ 13 in

Now that we know the height of the cylinder is about 13 in., we know the height of the cone, because the problem says that the height of the cone is half the height of the cylinder. 
half of 13 is 6.5, therefore the height of the cone is 6.5 
the radius of the cone is the same as that of the cylinder, 6 in. 

volume of a cone = πr² × (height ÷ 3)

volume of the cone = (3.14)(6²)(6.5 ÷ 3)
volume of the cone = (3.14)(36)(2.16666)
volume of the cone = 244.92 in³

Now all that's left to find the volume of the whole cake is to add the volume of the cylinder to the volume of the cone. 

1470 + 244.92 = 1714.92 in³



6 0
3 years ago
Plz plz help<br><br> Write a rule for the translation of AABC to AA'B'C'
Basile [38]
Don’t listen to people who give u links they are trying to scam you
6 0
3 years ago
Solve the system of equations.<br> y = 2x + 4<br> y = x2 + x - 2
miss Akunina [59]

Answer:

<u>x=6</u><u> y=16</u>

Step-by-step explanation:

I will use elimination

y=2x+4

y=2x+x-2

*3 y=2x+4 3y=6x+12

*2 y=3x-2 2y=6x-4

3y=6x+12

- 2y=6x. -4

<u>y=16</u>

16=2x+ 4

-4 -4

12= 2x

/2 /2

<u>6=x</u>

7 0
3 years ago
The author drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 199 times. Here are the observed
Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28​, 29​, 47​, 40​, 22​, 33.

Probability of an Event

      =\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911

Result is not significant.

3.

(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378

Result is not significant.

4.

(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349

Result is not significant.

5.

(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773

Result is not significant.

6.

(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair​ die.

   

8 0
2 years ago
I need help : D pleasee answer
Blizzard [7]
There’s nothing showing up. I think you didn’t add a picture
4 0
2 years ago
Read 2 more answers
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