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3 years ago

Whatdkdjfdjjfksakknshaja

Mathematics

1 answer:

trapecia [35]3 years ago

8 0

i think you have to multiply those. thats it

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69) Half-way to the center of a planet of uniform density, your weight compared to that at the surface would be :

mel-nik [20]

Answer:

One- half

Step-by-step explanation:

Since the density is uniform

Mass of the sphere = [(4/3) π r^3] d

where d is the uniform density.

Mass of the sphere = [(4/3) π d] r^3 = k r^3

where k = [(4/3) π d] is a constant

Weight = mg = G m M / r^2 = G m [k r^3] /r^2 = G m k r

Using Gauss’ law for gravitation,

Half way to the center of a planet the weight is only due to the inner sphere and the outer sphere does not contribute to his weight,

Inside his weight is mg’ = (G m k r) /2 = mg/2

Answer is one-half.

7 0

2 years ago

WHICH RATIO SHOW PORTION FORM WITH 25/28 BY WRITING IN LOWEST FORM

miss Akunina [59]

25/28

is in it's lowest form, as it cannot be reduced anymore.

25/28 = 0.8928571

hope this helps

8 0

3 years ago

What is the volume of a right circular cone if the circumference of the base is 125.6 ft and the height is 75 ft.? Use π=3.14 an

Alex_Xolod [135]

Answer:

The volume of a right circular cone is $31400\ \text{feet}^3$ .

Step-by-step explanation:

The circumference of the base of a right circular cone is 125.6 ft.

Height of cone is 75 ft.

Circumference of base is :

$C=2\pi r$ , r is radius

$r=\dfrac{C}{2\pi}\\\\r=\dfrac{125.6}{2\times 3.14}\\\\r=20\ ft$

The volume of a cone is given by :

$V=\dfrac{1}{3}\pi r^2 h\\\\V=\dfrac{1}{3}\times 3.14\times (20)^2\times 75\\\\V=31400\ \text{feet}^3$

So, the volume of a right circular cone is $31400\ \text{feet}^3$ .

5 0

2 years ago

Anyone know the answer to this ??

lana [24]

Answer: 23%
explanation:
4 + 6 + 2 + 2 + 3 + 4 + 6 + 3
There are total of 30 students.
4 + 3
7 students are freshman.
Find the percentage.
7/30 × 100
700/30 = 23.333333...
Round to nearest whole percent.
⇒ 23%

6 0

2 years ago

A recent survey based on a random sample of n=470 voters, predicted that independent candidate for the mayoral election will get

julsineya [31]

I have an expression

$\sigma = \sqrt{p(1-p)/n}$

floating around in my head; let's see if it makes sense.

The variance of binary valued random variable b that comes up 1 with probability p (so has mean p) is

$E( (b-p)^2 ) = (-p)^2(1-p) + (1-p)^2p = p(1-p)$

That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:

$\sigma = \sqrt{p(1-p)/n}$

Plugging in the numbers,

$\sigma = \sqrt{.24(1-.24)/490} = 0.019 = 1.9\%$

One standard deviation of the average is almost 2% so a 27% outcome was 3/1.9 = 1.6 standard deviations from the mean, corresponding to a two sided probability of a bit bigger than 10% of happening by chance.

So this is borderline suspect; most surveys will include a two sigma margin of error, say plus or minus 4 percent here, and the results were within those bounds.

7 0

3 years ago