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2 years ago

Determine the density in g/cm3 of a mineral that has a volume of 3.8 cm3 and a mass of 22.4 g.

Mathematics

1 answer:

kherson [118]2 years ago

4 0

Answer:

5.89 g/cm³

Step-by-step explanation:

givens,

Mass = 22.4 g

volume = 3.8 cm³

Density = mass / volume

= 22.4 / 3.8

= 5.89 g/cm³

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Which of the following statement is true about k-NN algorithm?

BaLLatris [955]

The answer is D. All of the above.

The computational complexity of K-NN increases as the size of the training data set increase and the algorithm gets significantly slower as the number of examples and independent variables increase.

Also, K-NN is a non-parametric machine learning algorithm and as such makes no assumption about the functional form of the problem at hand.

The algorithm works better with data of the same scale, hence normalizing the data prior to applying the algorithm is recommended.

6 0

2 years ago

A silicon chip is 14 nanometers thick. A nanometer is equal to 0.000000001 meter. Express the thickness of the chip using scient

oksano4ka [1.4K]

Answer:

1.4 × 10^-8

Step-by-step explanation:

The chip is 14 nanometers

14 * .000000001

.000000014

Move the decimal 8 places to the right, because we need 1 number in front of the decimal for scientific notation. The exponent will be -8 because we moved it 8 places to the right

1.4 × 10^-8

4 0

2 years ago

Read 2 more answers

The angle \theta_1θ

enyata [817]

Answer:

$sin\theta_1 = \dfrac{\sqrt{217}}{19}$

Step-by-step explanation:

It is given that:

$cos\theta_1 = -\dfrac{12}{19}$

And we have to find the value of $sin\theta_1 = ?$

As per trigonometric identities, the relation between $sin\theta\ and \ cos\theta$ can represented as:

$sin^2\theta + cos^2\theta = 1$

Putting $\theta_1$ in place of $\theta$ Because we are given

$sin^2\theta_1 + cos^2\theta_1 = 1$

Putting value of cosine:

$cos\theta_1 = -\dfrac{12}{19}$

$sin^2\theta_1 + (\dfrac{12}{19})^2 = 1\\\Rightarrow sin^2\theta_1 + \dfrac{144}{361} = 1\\\Rightarrow sin^2\theta_1 = 1-\dfrac{144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{217}{361}\\\Rightarrow sin\theta_1 = +\sqrt{\dfrac{217}{361}}, -\sqrt{\dfrac{217}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{\sqrt{217}}{19}, -\dfrac{\sqrt{217}}{19}$

It is given that $\theta_1$ is in 2nd quadrant and value of sine is always positive in 2nd quadrant. So, the answer is.

$\Rightarrow sin\theta_1 = \dfrac{\sqrt{217}}{19}$

8 0

2 years ago

ALGEBRA 2/ TRIG QUESTION ATTACHED Please help ❤️❤️

ss7ja [257]

Let's calculate the value of angle A and B

sin(A) =-4/5 → sin⁻¹(- 4/5) = A → A = - 53.13

cos(B) = -5/13 → cos⁻¹ (- 5/13) = B → B = 112.62

tan (A+B) = sin(A+B)/cos(A+B) with A+B = -53.13 + 112.62 = 59.49

tan (A+B) = sin(59.49)/cos(59.49) = 0.86154/0.507688 = 1.6969.

(Answer H = 56/33 = 1.6969)

7 0

2 years ago

(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX

pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable X follows a Normal distribution with mean μ = 10 and variance σ² = 36.

(a)

Compute the value of P (X > 5) as follows:

$P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z$

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

$P(4$

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

$P(X$

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

$P(X$

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

$P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z$

Thus, the value of P (X > 16) is 0.1587.

**Use a z-table for the probabilities.

8 0

2 years ago