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hodyreva [135]
3 years ago
8

6. The mass of one coin is 16.718 grams. The mass of a second coin is 27.22 grams. How much greater is the mass of the second co

in than the first? Show your reasoning.​
Mathematics
2 answers:
Alenkinab [10]3 years ago
4 0
The basss is really cool
lora16 [44]3 years ago
3 0

Answer:

The mass is greater by 10.502 grams. If you subtract 27.33 from 16.718, you get  10.502. Therefore, the mass of the second coin is greater than the first coin by 10.502 grams.

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grigory [225]

189 : 24 = 7 r 21

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3 years ago
-5÷(-3 1/3)= how do I solve this?​
liberstina [14]

Answer:

3/2 or 1.5

Step-by-step explanation:

Convert (-3 1/3) to -10/3

Now its -5/(-10/3)

The 3 is multiplied by the -5

(-5)(3)/(-10)

=(-15)/(-10)

The negatives cancel and it reduces to 3/2

8 0
3 years ago
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A sales team estimates that the number of new phones they will sell is a function of the price that they set. They estimate that
USPshnik [31]

Answer:

$204

Step-by-step explanation:

The question is at what price x will the company maximize revenue.

The revenue function is:

R(x) = 4,080x-10x^2

The price for which the derivate of the revenue function is zero is the price the maximizes revenue:

R(x) = 4,080x-10x^2\\\frac{dR(x)}{dx}=0=4,080-20x\\x=\frac{4,080}{20}\\ x=\$204

The company will maximize its revenue when the price is $204.

7 0
3 years ago
1000000000000 x 21392938402840
aivan3 [116]

Answer:

2.1392938e+25Step-by-step explanation: you just times it so u get this answer

4 0
2 years ago
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Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X > 10) b. P(X > 20) c
GrogVix [38]

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(c) The value of P (X < 30) is 0.9502.

(d) The value of x is 30.

Step-by-step explanation:

The probability density function of an exponential distribution is:

f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0

The value of E (X) is 10.

The parameter λ is:

\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10

(a)

Compute the value of P (X > 10) as follows:

P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679

Thus, the value of P (X > 10) is 0.3679.

(b)

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

P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353

Thus, the value of P (X > 20) is 0.1353.

(c)

Compute the value of P (X < 30) as follows:

P(X

Thus, the value of P (X < 30) is 0.9502.

(d)

It is given that, P (X < x) = 0.95.

Compute the value of <em>x</em> as follows:

P(X

Take natural log on both sides.

ln(e^{-0.10x})=ln(0.05)\\-0.10x=-2.996\\x=\frac{2.996}{0.10}\\ =29.96\\\approx30

Thus, the value of x is 30.

7 0
3 years ago
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