Need help checking my answers thanks!!

You have 42,784 grams of a radioactive kind of curium. If its half-life is 18 years, how much will be left after 72 years?

s344n2d4d5 [400]

Answer:

2,674.14 g

Step-by-step explanation:

Recall that the formula for radioactive decay is

N = N₀ e^(-λt)

where,

N is the amount left at time t

N₀ is the initial amount when t=0, (given as 42,784 g)

λ = coefficient of radioactive decay

= 0.693 ÷ Half Life

= 0.693 ÷ 18

= 0.0385

t = time elapsed (given as 72 years)

e = exponential constant ( approx 2.7183)

If we substitute these into our equation:

N = N₀ e^(-λt)

= (42,787) (2.7183)^[(-0.0385)(72)]

= (42,787) (2.7183)^(-2.7726)

= (42,787) (0.0625)

= 2,674.14 g

6 0

3 years ago

Evaluate g(x) = x2 + 5 given the inputs (-1,0,1,2}

mihalych1998 [28]

Answer:

g(-1) = 6

g(0) = 5

g(1) = 6

g(2) = 9

Step-by-step explanation:

Step 1: Plug in -1 for x

g(-1) = (-1)² + 5

g(-1) = 1 + 5

g(-1) = 6

Step 2: Plug in 0 for x

g(0) = 0² + 5

g(0) = 0 + 5

g(0) = 5

Step 3: Plug in 1 for x

g(1) = 1² + 5

g(1) = 1 + 5

g(1) = 6

Step 4: Plug in 2 for x

g(2) = 2² + 5

g(2) = 4 + 5

g(2) = 9

5 0

3 years ago

Can a Scalene triangle be a isosceles triangle? A: Always B: Never C: Sometimes

arsen [322]

A scalene triangle can never be an isosceles triangle. How? Well, let's take a look at the definition of a scalene triangle. In short, a scalene triangle is a classification of a triangle that does not have any congruent sides. On the other hand, an isosceles triangle is a triangle in which its legs are both congruent. Therefore, if a scalene triangle has two congruent legs, it wouldn't even be a scalene triangle, but instead it would be an isosceles triangle. Hope this helped!

3 0

3 years ago

G(t)= t^2 + 3 h(t) = 4t – 3 Find (g · h)(-1)

MArishka [77]

Answer:

fifty-two. Explanation: g(h(−1 ))=g(x). x=h(−1)=4(−1)−3=−4−3=−7. g(x)= g(−7)=72+3=49+3=52.

6 0

3 years ago

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 i

stich3 [128]

Complete question :

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 inches. Round answers to the nearest tenth of a percent. a) What percentage of years will have an annual rainfall of less than 43 inches? b) What percentage of years will have an annual rainfall of more than 39 inches? c) What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

Answer:

0.61053

0.66314

0.26647

Step-by-step explanation:

Given that :

Mean (m) = 41.4 inches

Standard deviation (s) = 5.7 inches

a) What percentage of years will have an annual rainfall of less than 43 inches?

P(x < 43)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (43 - 41.4) / 5.7 = 0.2807017

p(Z < 0.2807) = 0.61053 ( Z probability calculator)

b) What percentage of years will have an annual rainfall of more than 39 inches? c)

P(x > 39)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (39 - 41.4) / 5.7 = −0.421052

p(Z > −0.421052) = 0.66314 ( Z probability calculator)

What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

P(x < 38)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (38 - 41.4) / 5.7 = −0.596491

p(Z < −0.596491) = 0.27545 ( Z probability calculator)

P(x < 42)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (42 - 41.4) / 5.7 = 0.1052631

p(Z < 0.1052631) = 0.54192 ( Z probability calculator)

0.54192 - 0.27545 = 0.26647

6 0

3 years ago