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4 years ago

HELP WITH VOLUME OF SPHERE QUESTION!

Mathematics

1 answer:

Veronika [31]4 years ago

7 0

I found the volume for both spheres and subtracted the smaller volume from the larger and got 56972.16

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A system of equations has 1 solution. If 4x - y = 5 is one of the equations, which could be the other equation?

IrinaVladis [17]

Answer:

y = -4x+5

Step-by-step explanation:

I graphed the given equation on the graph below and graphed the choices with it. Only one choice had one solution with the given equation. This equation is y = -4x+5

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3 years ago

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Step-by-step explanation:

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3 years ago

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A batch of 330 samples of rejuvenated mitochondria contains 6 that are mutated (or defective). Two are selected from the batch,

Leno4ka [110]

Answer:

(a) Probability = 0.01520

(b) Probability = 0.00028

(c) Probability = 0.96391

Step-by-step explanation:

Total mitochondria = 330

Defective = 6

(a) Without replacement, if the first selection contains a defective mitochondria, we have a total of 5 defective remaining from a total of 329 mitochondria. Thus the probability the second one is defective Given the first one was defective will be :

Probability second is defective given the first one is not = 5 / 329 = 0.01520

(b) The probability the first is defective is : 6 / 330 = 0.01818

The probability the second is defective (without replacement) = 5 / 329 = 0.01520

The probability both are defective = 0.01818 * 0.01520 = 0.00028

(c) Probability the first one is acceptable = 324 / 330 = 0.98182

Probability the second one is acceptable = 323 / 329 = 0.98176

Probability both are acceptable = 0.98182 * 0.98176 = 0.96391

3 0

4 years ago

If (2i/2+i)-(3i/3+i)=a+bi, then a=<br>A. 1/10<br>B. -10<br>C. 1/50<br>D. -1/10

Verizon [17]

Answer:

Option A is correct.

Step-by-step explanation:

We are given:

$\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi$

We need to find the value of a.

The LCM of (2+i) and (3+i) is (2+i)(3+i)

$=\frac{2i(3+i)}{(2+i)(3+i)}-\frac{3i(2+i)}{(2+i)(3+i)}\\=\frac{6i+2i^2}{(2+i)(3+i)}-\frac{6i+3i^2}{(2+i)(3+i)}\\=\frac{6i+2i^2-(6i+3i^2)}{(2+i)(3+i)}\\=\frac{6i+2i^2-6i-3i^2)}{5+5i}\\=\frac{-i^2}{5+5i}\\i^2=-1\\=\frac{-(-1)}{5+5i}\\=\frac{1}{5+5i}$

Now rationalize the denominator by multiplying by 5-5i/5-5i

$=\frac{1}{5+5i}*\frac{5-5i}{5-5i} \\=\frac{5-5i}{(5+5i)(5-5i)}\\=\frac{5-5i}{(5+5i)(5-5i)}\\(a+b)(a-b)= a^2-b^2\\=\frac{5(1-i)}{(5)^2-(5i)^2}\\=\frac{5(1-i)}{25+25}\\=\frac{5(1-i)}{50}\\=\frac{1-i}{10}\\=\frac{1}{10}-\frac{i}{10}$