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galben [10]
3 years ago
6

The weight of a radioactive isotope was 96 grams at the start of an experiment. After one hour, the weight of the isotope was ha

lf of its initial weight. After two hours, the weight of the isotope was half of its weight the previous hour. If this pattern continues, which of the following graphs represents the weight of the radioactive isotope over time?

Mathematics
2 answers:
sergiy2304 [10]3 years ago
5 0

Answer:  Initial Weight = W₁ = 96

Change factor = 1/2 = 0.5

Step-by-step explanation:

zimovet [89]3 years ago
3 0
The starting weight of the radioactive isotope = 96 grams

Weight after one hour is half of the starting weight. So the weight of the radioactive isotope after 1 hour = 48 grams

After 2 hours the weight is half as compared to the weight after previous hour. So weight after 2 hours = 24 grams.

This means, after every hour the weight is being halved. The half life of radioactive isotope is one hour.

Since after every hour, the weight is being halved, the weight of the isotope can be modeled by an exponential equation.

So,

Initial Weight = W₁ = 96
Change factor = 1/2 = 0.5

The general equation of the sequence will be:

\\  \\ W_{t}=96(0.5)^{t}

Here t represents the number of hours. Using various values of t we can find the weight of the radioactive isotope at that time.

We can plot the sequence using the above equation. The graph is attached below.

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Joanne has 20 books on crafts and cooking. she has 6 more cookbooks than craft books. how many of each does she have?​
marin [14]

Answer:

Crafts books: 7

Cookbooks: 13

Step-by-step explanation:

Let x be the crafts book

Let y be the cookbooks

Solve for x:

  1. Plug x and y in: x + y = 20
  2. Re-write: y = x + 6
  3. Plug in x + 6 instead of y: x + x + 6 = 20
  4. Combine like terms: 2x + 6 = 20
  5. Subtract 6 from each side, so it now looks like this: 2x = 14
  6. Divide each side by 2 to cancel out the 2 next to x. It should now look like this: x = 7

Solve for y:

  1. Re-use an equation from above: y = x + 6
  2. Plug in the value of x: y = 7 + 6
  3. 7 + 6 = 13
  4. So, y = 13

I hope this helps!

5 0
3 years ago
If you could just give me answers for 5-d for easy points.
dalvyx [7]

Answer:

a) per year

Step-by-step explanation:

<em><u>Hope this helped! Have a nice day! Plz mark as brainliest!!!</u></em>

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4 0
2 years ago
Rectangle ABCD has a length of 16 centimeters and a width of 8 centimeters. What is the length of diagonal AC? Use Pythagorean t
pochemuha
A^2+ b^2 = c^2
16^2+ 8^2= c^2
256+64=c^2
320=c^2
square root for both side to get AC about 17.9 centimeters. Hope it help!
6 0
3 years ago
Use the remainder theorem to find P(2) for P(x) = x^4 - 3x^3 - 6x^2 + 9 .
kramer

Step-by-step explanation:

soln;

given; p(x) = x^4 - 3x^3 - 6x^2 + 9

using remainder theorem;

p(x) = p(a)

      = p(2)

      = (2)^4 - 3(2)^3 - 6(2)^2 + 9

      = -23

4 0
2 years ago
In estimating the mean score on a fitness exam, we use an original sample of size and a bootstrap distribution containing bootst
zmey [24]

Answer:

Option A is correct.

The confidence interval becomes

66 to 74.

Step-by-step explanation:

Complete Correct Question

In estimating the mean score on a fitness exam, we use an original sample of size n = 30 and a bootstrap distribution containing 5000 bootstrap samples to obtain a 95% confidence interval of 67 to 73. A change in this process is described below. If all else stays the same, which of the following confidence intervals (A, B, or C) is the most likely result after the change: Using an original sample of size n=16?A. 66 to 74

B. 67 to 73

C. 67.5 to 72.

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

To calculate the Sample Mean, we will use the confidence imterval obtained for the first sample size of 30 was 67 to 73.

Let the mean be x and the margin of error = y.

Lower limit of the confidence interval = x - y = 67

Upper limit of the confidence interval = x + y = 73

Solving simultaneously,

Sample mean = 70

Margin of error = 3

Then, since the sample size was initially 30, the critical value was obtained from the z-tables.

95% critical value = z = 1.960

Margin of error = (critical value) × (standard error)

3 = 1.960 × (standard error)

Standard error = 1.5306

Standard error of the mean = σₓ = (σ/√n) = 1.5306

σ = standard deviation of the sample = ?

n = sample size = 30

1.5306 = (σ/√30)

σ = 1.5306 × √30 = 8.3835

Note that using the central limit theorem, we can assume that sample mean for a sample size of 30 people will be approximately equal to the same value, digital value, that is Adam score of 70.

Sample mean for sample size 16, = 70

Sample standard deviation = 8.3835

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

Critical value = 1.960

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 8.3835

n = sample size = 16

σₓ = (8.3835/√16) = 2.096

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 70 ± (1.960 × 2.096)

CI = 70 ± 4.10

95% CI = (65.9, 74.1)

95% Confidence interval = (66, 74)

Hope this Helps!!!

3 0
3 years ago
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