A tumor is injected with 0.52 grams of Iodine-125. After 1 day, the amount of Iodine-125 has decreased by 1.15%.
Write an exponential decay model with A(t) representing the amount of Iodine-125 remaining in the tumor after t days. Enclose arguments of the function in parentheses and include a multiplication sign between terms. For example, c*ln(t).
Then use the formula for A(t) to find the amount of Iodine-125 that would remain in the tumor after 8.5 days.

Answers

Answer 1

Answer:

The equation is [tex]A(t) = 0.52*(0.9885)^{(t)}[/tex].

0.4713 grams would remain in the tumor after 8.5 days.

Step-by-step explanation:

Exponential equation of decay:

The exponential equation for the amount of a substance that decays, after t days, is given by:

[tex]A(t) = A(0)*(1-r)^t[/tex]

In which A(0) is the initial amount and r is the decay rate, as a decimal.

A tumor is injected with 0.52 grams of Iodine-125.

This means that [tex]A(0) = 0.52[/tex]

After 1 day, the amount of Iodine-125 has decreased by 1.15%.

This means that [tex]r = 0.0115[/tex]

So

[tex]A(t) = A(0)*(1-r)^t[/tex]

[tex]A(t) = 0.52*(1-0.0115)^t[/tex]

[tex]A(t) = 0.52*(0.9885)^t[/tex]

Then use the formula for A(t) to find the amount of Iodine-125 that would remain in the tumor after 8.5 days.

This is A(8.5).

[tex]A(t) = 0.52*(0.9885)^t[/tex]

[tex]A(8.5) = 0.52*(0.9885)^{8.5} = 0.4713[/tex]

0.4713 grams would remain in the tumor after 8.5 days.


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Clarissa wants to create more shade in her yard to block the hot sun, so she buys a sapling of an oak tree and plants it in the yard. To make sure the tree is growing at a healthy rate, she makes the following table to track its height over a period of weeks. What is the average rate of change in the tree's height from week 3 to week 9? Enter your solution into the box below.

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Answer:

The answer is 7.

Step-by-step explanation:

Determine what values to use from the table to find the average rate of change.  

The problem asks for the average rate of change (AROC) from week 3 to week 9. The points referring to these measurements would be (3, 41) and (9, 83), respectively.

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FInd the AROC from (3, 41) to (9, 83).

The formula for the slope of a line between two given points (which is the same as AROC from the first two the second point) is

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In your points, x1 is 3, x2 is 9, y1 is 41, and y2 is 83. Plug these values into the above formula and simplify.

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Step 3

Enter your solution.

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n 1990, the mean height of women 20 years of age or older was 63.7 inches based on data obtained from theCDC. Suppose that a random sample of 45 women who are 20 years of age or older in 2015 results in a meanheight of 63.9 inches with a standard deviation of 0.5 inch. Does your sample provide sufficient evidencethat women today are taller than in 1990

Answers

Answer:

The sample does not provide sufficient evidence that women today are taller than women in 1990.

The difference in mean height is not significant to arrive at such a conclusion.

Step-by-step explanation:

Mean height of women 20 year of age or older in 1990 = 63.7 inches

Mean height of women 20 years of age or older in 2015 = 63.9 inches

The standard deviation of the mean height in 2015 = 0.5 inch

The difference between the two years' mean height is 0.2 (63.9 - 63.7) inch.

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A rectangle is 17 inches long and 7 inches wide find the area

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Step-by-step explanation:

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Answer:

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Answer:

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Step-by-step explanation:

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use quadratic formula to get:

-11 ± √121 - 4(-1)/2

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* [tex]\sqrt{125}[/tex] = [tex]\sqrt{25}[/tex] · [tex]\sqrt{5}[/tex] = 5[tex]\sqrt{5}[/tex]

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Answer:

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Step-by-step explanation:

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A machine that is programmed to package 1.20 pounds of cereal is being tested for its accuracy In a sample of 36 cereal boxes, the sample mean filling weight is calculated as 1.22 pounds. The population standard deviation is known to be 0.06 pound. Identify the relevant parameter of interest for these quantitative data. The parameter of interest is the proportion filling weight for all cereal packages. The parameter of interest is the average filling weight for all cereal packages.
A-1. Identify the relevant parameter of interest for these quantitative data.
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B) The parameter of interest is the proportion filling weight of all cereal packages.
A-2. Compute its point estimate as well as the margin of error with 95% confidence. (Round intermediate calculations to 4 decimal places.
B-1. Calculate the 95% confidence interval.
B-2. Can we conclude that the packaging machine is operating improperly?
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B) Yes, since the confidence interval does not contain the target filling weight of 1.20.
C) No, since the confidence interval contains the target filling weight of 1.20.
D) No, since the confidence interval does not contain the target filling weight of 1.20.
C. How large a sample must we take if we want the margin of error to be at most 0.01 pound with 95% confidence?

Answers

Answer:

A) The parameter of interest is the average filling weight of all cereal packages. ;

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C) No, since the confidence interval contains the target filling weight of 1.20.

Step-by-step explanation:

A) The parameter of interest is the average filling weight of all cereal packages.

The point estimate = sample mean = 1.22

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Zcritical * σ/sqrt(n)

Zcritical at 95% confidence = 1.96

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Margin of Error : 1.96 * 0.01

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xbar ± margin of error

1.22 ± 0.0196

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(1.2004, 1.2396)

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Step-by-step explanation:

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Answers

Answer:

4/7

Step-by-step explanation:

The probability that a student does not play an instrument given that they play a sport is 4/23.

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It is the chance of an event to occur from a total number of outcomes.

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We have,

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Answer:

F

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In my backyard there are chickens and cows (no other animals). Among them there is a total of 366 eyes and 558 legs. How many chickens do I have in my backyard?

Answers

The total animals = 139, and we can substitute this value into equation (3) to obtain:x = 2(139) - 279 => x = -1This doesn't make sense since we can't have negative chickens.

Let the number of chickens be x and the number of cows be y in the backyard. We can create two equations based on the given information:The total number of animals is x + y = total animals ...

(1)The total number of eyes and legs can also be expressed as 2x + 2y = 366 (since each animal has two eyes) and 2x + 4y = 558 (since each chicken has 2 legs and each cow has 4 legs).

Simplifying the equations, we get:x + y = total animals ... (1)x + 2y = 279 ... (2)From equation (1),

we can substitute y = total animals - x in equation (2) to obtain:x + 2(total animals - x) = 279 => 2total animals - x = 279 => x = 2total animals - 279 ... (3)We also know that x + y = total animals, so y = total animals - x.

Substituting equation (3) into this expression, we get:y = total animals - (2total animals - 279) => y = 279 - total animals ... (4)

To find the value of x, we need to know the value of total animals. We can obtain this value by adding equations (1) and (4):x + y + (279 - total animals) = total animals => x + y = 279/2 => x + y = 139.5

Since x and y are both whole numbers, they must add up to an even number. The only possible value is 70 + 69 = 139.  Therefore, there must be a mistake in the problem statement or calculations.

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A sample size 25 is picked up at random from a population which is normally
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Answers

Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 100 and variance of 36.

This means that [tex]\mu = 100, \sigma = \sqrt{36} = 6[/tex]

Sample of 25:

This means that [tex]n = 25, s = \frac{6}{\sqrt{25}} = 1.2[/tex]

(a) P(X<99)

This is the pvalue of Z when X = 99. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{99 - 100}{1.2}[/tex]

[tex]Z = -0.83[/tex]

[tex]Z = -0.83[/tex] has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{100 - 100}{1.2}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a pvalue of 0.5

X = 98

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{98 - 100}{1.2}[/tex]

[tex]Z = -1.67[/tex]

[tex]Z = -1.67[/tex] has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

So

P(98 < X < 100) = 0.4525


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Answers

It C)
This is because the sum of the two smaller sides in a triangle has to be larger than the bigger sides. 10+12=22.
22 is smaller than 24 which is the third side this no triangle is formed

This had me confused i need big help!!!!

Answers

Answer:

[tex]3\frac{7}{8}[/tex]

Step-by-step explanation:

Equation: 4 1/7 + 2 1/8 + x= 10 1/7

Convert into improper fractions:

29/7 + 17/8= 71/7

Add fractions

351/56+x= 71/7

Subtract on both sides

X= 31/8 or 3 7/8

Name the figure below in two different ways.

Answers

Answer:

LDM

MDL

Step-by-step explanation:

The names of a figure are gotten from the points on the figure. The names of the figure are LDM  and MDL.

There are three points on the given figure, and these points are: point L, point D and point M, where Point D is between L and M.

This means that, when naming the figure, alphabet D must be at the middle while alphabets M and L can be at either sides of D.

So, the possible names of the figure are: LDM  and MDL.

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