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3 years ago

10

I need help on this math

1 answer:

yanalaym [24]3 years ago

4 0

G (5)=27, because it shows it’s in the same table

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4.<br> Find the geometric mean of 4 and 12.<br><br><br> 24<br><br><br><br><br><br><br><br><br> 8

r-ruslan [8.4K]

The geometric mean of two numbers is the square root of their product.

sqrt{4 • 12}

sqrt{48}

sqrt{16} •sqrt{3}

4•sqrt{3}.

The geometric mean of 4 and 12 is

4•sqrt{3}.

5 0

3 years ago

Nilton acude a una institución privada que ofrece una tasa de interés del 0,5% mensual por el dinero que deposites, a un plazo f

jeka57 [31]

Answer:

hey, i might help in english?

Step-by-step explanation:

3 0

3 years ago

Modeling Radioactive Decay In Exercise, complete the table for each radioactive isotope.

Travka [436]

Answer :

The amount after 1000 years will be, 5.19 grams.

The amount after 10000 years will be, 0.105 grams.

Step-by-step explanation :

Half-life = 1599 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{1599\text{ years}}$

$k=4.33\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the amount after 1000 years.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $4.33\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = 1000 years

a = initial amount of the reactant = 8 g

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$1000=\frac{2.303}{4.33\times 10^{-4}}\log\frac{8}{a-x}$

$a-x=5.19g$

Thus, the amount after 1000 years will be, 5.19 grams.

Now we have to calculate the amount after 10000 years.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $4.33\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = 10000 years

a = initial amount of the reactant = 8 g

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$10000=\frac{2.303}{4.33\times 10^{-4}}\log\frac{8}{a-x}$

$a-x=0.105g$

Thus, the amount after 10000 years will be, 0.105 grams.

4 0

3 years ago

(Anderson, 1.14) Assume that P(A) = 0.4 and P(B) = 0.7. Making no further assumptions on A and B, show that P(A ∩ B) satisfies 0

Goshia [24]

Answer with Step-by-step explanation:

We are given that

P(A)=0.4 and P(B)=0.7

We know that

$P(A)+P(B)+P(A\cap B)=P(A\cup B)$

We know that

Maximum value of $P(A\cup B)$ =1 and minimum value of $P(A\cup B)$ =0

$0\leq P(A\cup B )\leq 1$

$0\leq P(A)+P(B)-P(A\cap B)\leq 1$

$0\leq 0.4+0.7-P(A\cap B)\leq 1$

$0\leq 1.1-P(A\cap B)\leq 1$

$0\leq 1.1-P(A\cap B)$

$P(A\cap B)\leq 1.1$

It is not possible that $P(A\cap B)$ is equal to 1.1

$1.1-P(A\cap B)\leq 1$

$-P(A\cap B)\leq 1-1.1=-0.1$

Multiply by (-1) on both sides

$P(A\cap B)\geq 0.1$

Again, $P(A\cup B)\geq P(B)$

$0.4+0.7-P(A\cap B)\geq 0.7$

$1.1-P(A\cap B)\geq 0.7$

$-P(A\cap B)\geq -1.1+0.7=-0.4$

Multiply by (-1) on both sides

$P(A\cap B)\leq 0.4$

Hence, $0.1\leq P(A\cap B)\leq 0.4$

3 0

3 years ago

Please answer #1 and #2

DIA [1.3K]

1)
x^2 + (3y/2z) = 7
2x^2z + 3y = 14z
3y = 14z - 2x^2z
3y = 2z(7 - x^2)
y = 2/3(z)(7 - x^2)

2)
(3zx^4) /(5+z) = 2y
3zx^4 = 2y(5+z)
3zx^4 = 10y + 2yz

6 0

3 years ago