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3 years ago

Hiii I need help in this Plzzz helppp

Mathematics

2 answers:

Pepsi [2]3 years ago

5 0

To round to the nearest hundredth, look at the third value to the right of the decimal point.

For instance, if there is a value 7.829, it would be rounded to 7.83, because 9 rounds up to 10. If the value was instead 7.821, it would be rounded to 7.82, because 1 rounds to 0.

However, the values listed are already rounded to the nearest hundredth, so no more work is required.

Mumz [18]3 years ago

4 0

Answer:

No rounded needed.

Step-by-step explanation:

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Step-by-step explanation:

the answer is 52 because all the angles have to add up to 360 and 128+126+54=308 and 360-308=52

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A pattern has 8 blue triangles to every 8 yellow triangles. What is the ratio of yellow triangles to blue triangles?

Tpy6a [65]

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1:1

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if there are 8 of the yellow to 8 of the blue then the ratio would be 8:8 and that can simplify to 1:1 :)

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4 years ago

A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th

QveST [7]

we know that

the volume of a cone is equal to

$V= \frac{1}{3} \pi r^{2}h$

in this problem

the radius is equal to

$r= \frac{112}{2}= 56ft$

1) Find the height of the cone before the storm

Applying the Pythagorean Theorem find the height

$h^{2} = l^{2}-r^{2}$

$l=65 ft$

$h^{2} = 65^{2}-56^{2}$

$h^{2} = 1,089$

$h=33 ft$

2) Find the volume before the storm

$V= \frac{1}{3}*\pi* 56^{2}*33$

$V=34,496\pi\ ft^{3}$

3) Find the volume after the storm

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

r=(56/3) ft

h=(33/3)=11 ft

$V= \frac{1}{3}*\pi* (56/3)^{2}*11$

$V= 1,277.63\pi\ ft^{3}$

4) Find how this change affect the volume of the pile

Divide the volume after the storm by the volume before the storm

$\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}$

therefore

the answer part a) is

The volume of the pile after the storm is $\frac{1}{27}$ times the original volume

Part b) Estimate the number of lane miles that were covered with salt

5) Find the amount of salt that was used during the storm

$=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}$

6) Find the pounds of road salt used

$104,358.59*80=8,348,687.2\ pounds$

7) Find the number of lane miles that were covered with salt

$8,348,687.2/350=23,853.39 \ lane\ miles$

therefore

the answer part b) is

the number of lane miles that were covered with salt is $23,853.39 \ lane\ miles$

Part c) How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to $1,277.63\pi\ ft^{3}$

$1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}$

8) Find the pounds of road salt

$4,013.79*80=321,103.20\ pounds$

9) Find the number of lane miles

$321,103.20/350=917.44 \ lane\ miles$

therefore

the answer part c) is

the number of lane miles is $917 \ lane\ miles$

7 0

4 years ago

Why is this binomial and what is the answer?

pashok25 [27]

Answer:

2e⁻² ≈ 0.271

Step-by-step explanation:

This is actually a Poisson distribution. The probability is:

P(x; μ) = e^(-μ) (μˣ) / x!

P(2; 2) = e⁻² (2²) / 2!

P(2; 2) = 2e⁻²

P(2; 2) ≈ 0.271

5 0

4 years ago