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2 years ago

Tabitha defines a sphere as the set of all points equidistant from a single point. Is Tabitha's definition valid?

Mathematics

2 answers:

Advocard [28]2 years ago

5 0

definition is valid since center is exactly 1radius away from any point on the outersurface.

other figures do not fit this defn

Ipatiy [6.2K]2 years ago

4 0

Answer:

<h2>Tabitha's defintion is valid.</h2>

Step-by-step explanation:

Her definition is valid because the sphere is basically defined by its radius, which is the equidistant points towards a center. If points are not equidistant, then that's not a sphere.

Also, Tabitha is considering "a set of points" referring to infinite points on the surface of the sphere. The single point she mentioned is the center.

Therefore, Tabitha's definition is valid because she's considering all elements that define a sphere.

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If (3x2 + 22x + 7) ÷(x + 7) = 3x + 1, then (x + 7)( ) = . The check of the polynomial division problem shows that the product of

azamat

Answer:

$(x + 7) * (3x + 1) = (3x^2 + 22x + 7)$

Step-by-step explanation:

Given

$(3x^2 + 22x + 7) \div (x + 7) = 3x + 1$

$(x + 7) * (\ ) = [\ ]$

Required

Complete the blanks

We have:

$(3x^2 + 22x + 7) \div (x + 7) = 3x + 1$

Rewrite as:

$\frac{(3x^2 + 22x + 7) }{ (x + 7)} = 3x + 1$

Cross multiply

$(x + 7) * (3x + 1) = (3x^2 + 22x + 7)$

8 0

2 years ago

Read 2 more answers

Find the simple interest paid to the nearest cent for each loan amount, interest rate, and time.

myrzilka [38]

Answer:

I = $ 1,937.50

Equation:

I = Prt

Calculation:

First, converting R percent to r a decimal

r = R/100 = 3.875%/100 = 0.03875 per year,

then, solving our equation

I = 10000 × 0.03875 × 5 = 1937.5

I = $ 1,937.50

The simple interest accumulated

on a principal of $ 10,000.00

at a rate of 3.875% per year

for 5 years is $ 1,937.50.

Step-by-step explanation:

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2 years ago

The national average sat score (for verbal and math) is 1028. if we assume a normal distribution with standard deviation 92, wha

elena55 [62]

Let X be the national sat score. X follows normal distribution with mean μ =1028, standard deviation σ = 92

The 90th percentile score is nothing but the x value for which area below x is 90%.

To find 90th percentile we will find find z score such that probability below z is 0.9

P(Z <z) = 0.9

Using excel function to find z score corresponding to probability 0.9 is

z = NORM.S.INV(0.9) = 1.28

z =1.28

Now convert z score into x value using the formula

x = z *σ + μ

x = 1.28 * 92 + 1028

x = 1145.76

The 90th percentile score value is 1145.76

The probability that randomly selected score exceeds 1200 is

P(X > 1200)

Z score corresponding to x=1200 is

z = $\frac{x - mean}{standard deviation}$

z = $\frac{1200-1028}{92}$

z = 1.8695 ~ 1.87

P(Z > 1.87 ) = 1 - P(Z < 1.87)

Using z-score table to find probability z < 1.87

P(Z < 1.87) = 0.9693

P(Z > 1.87) = 1 - 0.9693

P(Z > 1.87) = 0.0307

The probability that a randomly selected score exceeds 1200 is 0.0307

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3 years ago

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PSYCHO15rus [73]

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2 years ago

Urgent!!!!!!!plzz

Nitella [24]

Answer:

too much reading, can you put more points.

(edit: OML SO SORRY -THOUGHT THIS WAS QUESTIONS)

Step-by-step explanation:

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3 years ago