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2 years ago

Find the half life of an element which decays by 3.409% each day

Mathematics

1 answer:

vodomira [7]2 years ago

7 0

Answer:

The half-life of a radioactive isotope is 140 How many days would it take for the decay rate of a sample of this isotope to fall to one-fourth of its initial value?

Step-by-step explanation:

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The midpoint of chord AB is ( , ), and the slope of chord AB is

kykrilka [37]

Answer:

See solution below

Step-by-step explanation:

Let the coordinate's of A and B be (1, 0) and (2,4) respectively

midpoint M (X, Y) = [(x1+x2/2, y1+y2/2)]

X = x1+x2/2

X = 1+2/2

X = 3/2

X = 1.5

Y = y1+y2/2

Y = 0+4/2

Y = 4/2

Y = 2

Hence the required midpoint (X, Y) is (1.5, 2)

Slope m = y2-y1/x2-x1

m = 4-0/2-1

m = 4/1

m = 4

Hence the slope is 4

<em>Note that the coordinates are assumed but the same calculation can be employed for any other coordinates</em>

7 0

2 years ago

What is the value of x? x+45=11 Enter your answer in the box in simplest form.

Ksivusya [100]

Isolate the x. Note the equal sign. What you do to one side, you do to the other. Subtract 45 from both sides

x + 45 (-45) = 11 (-45)

x = 11 - 45

x = -34

-34 is your answer

hope this helps

8 0

2 years ago

Read 2 more answers

A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between

sattari [20]

Answer:

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) 0.09

Step-by-step explanation:

We are given the following in the question:

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)$

where B(t) gives the brightness of the star at time t, where t is measured in days.

a) rate of change of the brightness after t days.

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(t) = 0.45\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\times \dfrac{2\pi}{4.4}\\\\B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) rate of increase after one day.

We put t = 1

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(1) = \dfrac{0.9\pi}{4.4}\bigg(\cos(\dfrac{2\pi (1)}{4.4}\bigg)\\\\B'(t) = 0.09145\\B'(t) \approx 0.09$

The rate of increase after 1 day is 0.09

8 0

2 years ago

Frank, who lives in Texas, and his sister Lilly, who lives in Japan, correspond regularly. From what he can tell from the postma

Hoochie [10]

Question:

The available options are:

(A) There is convincing evidence that there is no difference in the mean delivery times.

(B) There is convincing evidence that there is a difference in the mean delivery times.

(C) There is convincing evidence that the mean delivery time from Japan to Texas is greater than the mean delivery time from Texas to Japan.

(D) There is not convincing evidence that the mean delivery time from Japan to Texas is greater than the mean delivery time from Texas to Japan. (E) The t-test cannot be used for sample sizes that are this small.

Answer:

The correct option is;

(A) There is convincing evidence that there is no difference in the mean delivery times.

Step-by-step explanation:

Here we have the formula for the two-sample t-test given s;

$t =\frac{\overbar\overline{\rm x}_1 - \overbar\overline{\rm x}_2 }{\sqrt{\frac{\sigma ^2_1}{n_1} } +\frac{\sigma ^2_2}{n_2} }}$

To Texas N₁ = 12, $\overline{\rm x}_1$ =8.74 σ₁ = 2.92 SE Mean₁ = 0.84

To Japan N₂ = 9 $\overline{\rm x}_2$ = 6.75 σ₂ = 2.56 SE Mean₁ = 0.85

t = 1.667

Where the null hypothesis is

mu To Texas = mu To Japan and the alternative is

mu To Texas > mu To Japan

Here since the observed P value of P = 0.058 is greater than the significance level of 0.05 we fail to reject the null hypothesis and we will therefore, not accept the alternative hypothesis.

Hence, there is convincing evidence that there is no difference in the mean delivery times.

3 0

3 years ago

15 3/4 is less than -2+x Answers A. Add 15 3/4 to both sides of the inequality B.Subtract 2 from both sides of the inequality C.

DochEvi [55]

Answer:

Step-by-step explanation:

15 3/4 < -2+x

15 3/4 + 2< 2-2+x

71/4<x

5 0

3 years ago