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2 years ago

A sphere has a volume of 1,436. 03m3, what is the approximate length of its radius?.

Mathematics

1 answer:

makkiz [27]2 years ago

3 0

The sphere with the volume 1436.03 m³ has 7m long radius.

To find out the approximate length of its radius by using the formula to calculate volume of sphere is:

V= 4πr³/3

The value of π is 22/7 or 3.14.

So, V = 4 × 3.14 × r³ / 3

Volume of the sphere is given which is 1436.03 m³.

Thus, r³ = 1436.03×3 / 4×3.14

r = ³√ 4308.09/12.56 = ³√343 = 7 m.

Hence, the radius of the given sphere is 7 m.

The cube root of a number is the factor that we multiply by itself three times to get that number.

For example, the cube root of 27 is 3.

To learn more about the sphere click here brainly.com/question/19868745

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What are all the zeros of the polynomial y= -x^3 -3x^2 +6x +8 In synthetic division

crimeas [40]

Answer:

x=2, -4, -1

Step-by-step explanation:

-x³-3x²+6x+8=0

Multiply entire equation by -1:

x³+3x²-6x-8=0

Since x=2 is a solution, then x-2 must be a factor. Then:

x³+3x²-6x-8/x-2=x²+5x+4

x²+5x+4=0

Factor this:

x²+5x+4=(x+4)(x+1)=0

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x=2, -4, -1

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3 years ago

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3 years ago

A jar contains 2 black marbles and 1 red marble. Lisa randomly selects one marble, notes its color and then places it back in th

nasty-shy [4]

Answer :

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3 years ago

Find a polynomial of the form

Sloan [31]

Answer:

$f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3$

Step-by-step explanation:

We are given that

$f(x)=ax^3+bx^2+cx+d$

f(0)=-3,f(1)=2,f(3)=5 and f(4)=0

We have to find the polynomial

Substitute the value x=0 then ,we get

f(0)=d=-3

Substitute x=1 then we get

$a+b+c-3=2$

$a+b+c=2+3=5$

$a+b+c=5$ (equation I)

Substitute x=3 then we get

$a(3)^3+b(3)^2+c(3)-3=5$

$27a+9b+3c=5+3=8$

$27a+9b+3c=8$ (Equation II)

Substitute x=4 then we get

$a(4)^3+b(4)^2+c(4)-3=0$

$64a+16b+4c=3$ (Equation III)

Equation I multiply by 3 then subtract from equation II

$24a+6b=-7$ (Equation IV)

Equation II multiply by 4 and equation III multiply by 3 and subtract equation II from III

$84a+12b=-23$ (Equation V)

Equation IV multiply by 2 and then subtract from equation V

$36a=-9$

$a=-\frac{9}{36}$

$a=-\frac{1}{4}$

Substitute the value of a in equation IV then we get

$24(-\frac{1}{4})+6b=-7$

$-6+6b=-7$

$6b=-7+6=-1$

$b=-\frac{1}{6}$

Substitute the value of b in equation I then we get

$-\frac{1}{4}-\frac{1}{6}+ c=5$

$-\frac{5}{12}+c=5$

$c=5+\frac{5}{12}=\frac{65}{12}$

Substitute the values then we get

$f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3$

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4 years ago

Assuming the probability of a single sample testing positive is 0.15, find the probability of a positive result for two samples

Grace [21]

Answer:

$P(Positive\ Mixture) = 0.2775$

The probability is not low

Step-by-step explanation:

Given

$P(Single\ Positive) = 0.15$

$n = 2$

Required

$P(Positive\ Mixture)$

First, we calculate the probability of single negative using the complement rule

$P(Single\ Negative) = 1 - P(Single\ Positive)$

$P(Single\ Negative) = 1 - 0.15$

$P(Single\ Negative) = 0.85$

$P(Positive\ Mixture)$ is calculated using:

$P(Positive\ Mixture) = 1 - P(All\ Negative)$ ---- i.e. complement rule

So, we have:

$P(Positive\ Mixture) = 1 - 0.85^2$

$P(Positive\ Mixture) = 1 - 0.7225$

$P(Positive\ Mixture) = 0.2775$

Probabilities less than 0.05 are considered low.

So, we can consider that the probability is not low because 0.2775 > 0.05

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3 years ago