The probability that the sample mean would be greater than 133.5 is 0.9926
Explanation:Given:
Mean = 137
Standard deviation = 8
Sample = 31
To find the probability that the sample mean would be greater than 133.5, we have:
[tex]\begin{gathered} z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}} \\ \\ =\frac{133.5-137}{\frac{8}{\sqrt{31}}}=-2.4359 \\ \\ P(X>133.5)=1-P(z<-2.4359) \\ =1-0.0074274 \\ \approx0.9926 \end{gathered}[/tex]The number of visits to public libraries increased from 1.3 billion in 1999 to 1.5 billion in 2004. Find the average rate of change in the number of public library visits from 1999 to 2004.The average rate of change between 1999 and 2004 was: billion: Simplify your answer. Type an integer or a decimal.)
The average rate of change is defined as:
[tex]\frac{f(b)-f(a)}{b-a}[/tex]using the information given
a=1999
b=2004
f(a)=1.3
f(b)=1.5
then,
[tex]\begin{gathered} \frac{1.5-1.3}{2004-1999} \\ \frac{0.2}{5} \\ 0.04 \end{gathered}[/tex]The average rate of change between 1999 and 2004 was 0.04 billion.
I need help with the question
B
For this problem Let's work in parts
1) Coin
Heads
Tails
Flipping the coin once, the Probability is:
[tex]P\text{ =}\frac{1}{2}[/tex]For there are two possible results, Heads or Tails, and there was one flipping.
2) Spinner
1 to 6 sections
The Probability of this spinner lands on a number lesser than 3
[tex]P\text{ =}\frac{2}{6}\text{ = }\frac{1}{3}[/tex]is 1 out of 3 for this spinner, since only 1, 2 are valid results.
So, the answer to this experiment
[tex]P\text{ = }\frac{1}{3}\cdot\frac{1}{2}\text{ = }\frac{1}{6}[/tex]Is the probability of both happen, both spinner and coin are 1 in six flipping. Since there are only two numbers < 3 on the spinner and two possibilities for the coin.
B
Heads, 1
Tails 1
Heads2
Tails 2
X1 2 Given f(x) = 35 - 2 - 2
Use the rule of correspondence of the case when x>3, since 5>3.
[tex]\begin{gathered} f(5)=5+2 \\ =7 \end{gathered}[/tex]Therefore, f(5)=7.
Multiples of 36 and the square root of 49
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
36
√49
Step 02:
multiples:
36:
The multiples of thirty-six are the numbers that contain it a whole number of times.
36, 72, 108, 144, 180, ...
√49:
[tex]\sqrt[]{49}=7[/tex]The multiples of seven are the numbers that contain it a whole number of times.
7, 14, 21, 28, 35, 42, ...
That is the full solution.
Convert 5 1/4 lb to oz.
The conversion factor for lb to oz is
[tex]1lb=16oz[/tex]I will put the pounds first in terms of the improper fraction. We have
[tex]5\frac{1}{4}=\frac{21}{4}[/tex]Using the conversion factor to convert lb to oz, we have
[tex]\frac{21}{4}lb\times\frac{16oz}{1lb}=\frac{21\cdot16}{4}=\frac{21\cdot4}{1}=84oz[/tex]Hence, 5 1/4 lb is equal to 84 oz.
Answer: 84 oz
Using the compound interest formula, determine the total amount paid back and the monthly payment. Buying a $6000 used sedan taken out with $500 paid up front and the rest borrowed at 8.3%annual interest compounded daily (365 days per year) over 2 years.
The final value of an investment or loan with compound interest is given by:
[tex]FV=P(1+\frac{r}{m})^{m\cdot t}[/tex]Where P is the initial value (principal or loan), r is the annual interest rate, t is the duration of the investment/loan, and m is the number of compounding periods per year.
The following values are given in the problem:
P = $6000 - $500 = $5500
r = 8.3% = 0.083
t = 2 years
m = 365
Applying the formula:
[tex]FV=5500(1+\frac{0.083}{365})^{365\cdot2}[/tex]Calculating:
[tex]FV=5500(1+0.0002273926)^{730}[/tex]FV = $6493.03
The total amount paid back is $6493.03
This is equivalent to an approximate monthly payment of:
[tex]R=\frac{$ 6493.03 $}{24}=270.54[/tex]The monthly payment is approximately $270.54
If the sample space, S = {1, 2, 3, 4, …, 15} and A = the set of odd numbers from the given sample space, find Ac.A.{1, 2, 3, 4, 5, 6, …, 15}B.{1, 3, 5, 7, 9, 11, 13, 15}C.{1, 2, 3, 4, 15}D.{2, 4, 6, 8, 10, 12, 14}
A^c is the complement of set A.
Given that A is a subset of S, then A^c contains the elements present in set S but not in set A.
The sets are:
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
A = {1, 3, 5, 7, 9, 11, 13, 15} (odd numbers present in S)
Therefore, the elements present in set S but not in set A are:
[tex]A^c=\mleft\lbrace2,4,6,8,10,12,14\mright\rbrace[/tex]
For a standard normal distribution, find the z-value that goes with a left tail area=0.9931
The z-value that goes with a left tail Area= 0.9931 is 2.4 .
What is normal distribution?A probability distribution that is symmetric about the mean is the normal distribution, also known as the Gaussian distribution. Data close to the mean are more common than data far from the mean. The normal distribution is displayed as a "bell curve" on the chart.
What is left rear area?
The area under the curve to the left of x* in Figure 5.19, “Right and left tails of the distribution” is known as the left tail of the density curve for a continuous random variable X whose limit is x* (a).
According to the Z-value normal distribution table, its value is 2.4 .
To know more about normal distribution visit to:
https://brainly.com/question/13759327
#SPJ1
Apply zero product theorem to solve for x[tex]x ^{2} = 9[/tex]
Answer:
[tex]\begin{gathered} x_1=-3 \\ x_2=3 \end{gathered}[/tex]Step-by-step explanation:
To apply the zero product theorem, put all the terms on the left side to equal zero.
[tex]x^2-9=0[/tex]Factoring the binomial:
[tex]\begin{gathered} (x+3)(x-3)=0 \\ x_1+3=0 \\ x_1=-3 \\ \\ x_2-3=0 \\ x_2=3 \end{gathered}[/tex]-82638•9390(69)+420 please help me with this
EXPLANATION
Given the operation -82638•9390(69)+420, multiplying numbers and applying the sign rule:
=-775970820(60) + 420
Applying the distributive property:
= -46558249200 + 420
Adding numbers:
= -46558248780
The solution is -46558248780
please see the picture below. I'll only need b c and d
Given:
• cotθ = -3
,• secθ < 0
,• 0 ≤ θ < 2π
Here the cot value of the angle is negative.
The cotangent function is negative in quadrants II and IV.
Also, secθ < 0, which means it is negative.
Secant function is negative in II and III quadrants.
Therefore, the angle will be in quadrant II.
Let's find the exact values of the following:
• (a). sin(2θ)
Apply the double angle formula:
[tex]sin(2\theta)=2sin\theta cos\theta=\frac{2tan\theta}{1+tan^2\theta}[/tex]Where:
[tex]tan\theta=\frac{1}{cot\theta}=-\frac{1}{3}[/tex]Thus, we have:
[tex]\begin{gathered} sin(2\theta)=\frac{2*(-\frac{1}{3})}{1+(-\frac{1}{3})^2} \\ \\ sin(2\theta)=\frac{-\frac{2}{3}}{1+\frac{1}{9}}=\frac{-\frac{2}{3}}{\frac{9+1}{9}}=\frac{-\frac{2}{3}}{\frac{10}{9}} \\ \\ sin(2\theta)=-\frac{2}{3}*\frac{9}{10} \\ \\ sin(2\theta)=-\frac{3}{5} \\ \\ \text{ Sine is positive in quadrant II:} \\ sin(2\theta)=\frac{3}{5} \end{gathered}[/tex]• cos(2θ):
Apply the formula:
[tex]cos(2\theta)=\frac{1-tan^2\theta}{1+tan^2\theta}[/tex]Thus, we have:
[tex]\begin{gathered} cos(2\theta)=\frac{1-(-\frac{1}{3})^2}{1+(-\frac{1}{3})^2} \\ \\ cos(2\theta)=\frac{1-\frac{1}{9}}{1+\frac{1}{9}} \\ \\ cos(2\theta)=\frac{\frac{9-1}{9}}{\frac{9+1}{9}}=\frac{\frac{8}{9}}{\frac{10}{9}}=\frac{8}{9}*\frac{9}{10}=\frac{4}{5} \\ \\ cos(2\theta)=\frac{4}{5} \\ \text{ } \\ \text{ Cosine is negative in quadrant II>} \\ cosine(2\theta)=-\frac{4}{5} \end{gathered}[/tex]• (c). sin(θ/2):
Apply the formula:
[tex]cos\theta=1-2sin^2(\frac{\theta}{2})[/tex]Where:
[tex]tan\theta=\frac{opposite}{adjacent}=-\frac{1}{3}[/tex]Now, let's find the hypotenuse using Pythagorean Theorem:
[tex]\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10}[/tex]Thus, we have:
[tex]cos\theta=\frac{adjacent}{hypotenuse}=-\frac{3}{\sqrt{10}}[/tex]Now, the function will be:
[tex]\begin{gathered} cos\theta=1-2sin^2(\frac{\theta}{2}) \\ \\ -\frac{3}{\sqrt{10}}=1-2sin^2(\frac{\theta}{2}) \\ \\ 2sin^2(\frac{\theta}{2})=1+\frac{3}{\sqrt{10}} \\ \\ 2sin^2(\frac{\theta}{2})=\frac{10+3\sqrt{10}}{10} \\ \\ sin^2(\frac{\theta}{2})=\frac{10+3\sqrt{10}}{20} \\ \\ sin(\frac{\theta}{2})=\sqrt{\frac{10+3\sqrt{10}}{20}} \end{gathered}[/tex]• (d). cos(,(θ/2)):
[tex]\begin{gathered} 2cos\theta=2cos^2(\frac{\theta}{2})-1 \\ \\ cos\frac{\theta}{2}=\sqrt{\frac{1+cos\theta}{2}}=\sqrt{\frac{1-\frac{3}{\sqrt{10}}}{2}} \end{gathered}[/tex]ANSWER:
[tex]\begin{gathered} (a).\text{ }\frac{3}{5} \\ \\ \\ (b).\text{ -}\frac{4}{5} \\ \\ \\ (c).\text{ }\sqrt{\frac{10+3\sqrt{10}}{20}} \\ \\ \\ (d).\text{ }\sqrt{\frac{1-\frac{3}{\sqrt{10}}}{2}} \end{gathered}[/tex]All changes 4. What are the coordinates of the midpoint of the line segment with endpoints (7, 2) and (3, 4)? O (5,3) O (4, -2) O (4,2) 0 (2, 1)
We will find the coordinates of the mid-point using the following expression:
[tex]mp=(\frac{_{}x_2+x_1}{2},\frac{y_2+y_1}{2})_{}_{}_{}[/tex]So, when we replace we obtain the mid-point coordinates:
[tex]mp=(\frac{7+3}{2},\frac{2+4}{2})\Rightarrow mp=(5,3)[/tex]So, the coordinates of the mid-point are (5, 3).
what is 39 ÷ (2+ 1) - 2 × (4 + 1)
The given expression is
[tex]36\colon(2+1)-2\times(4+1)[/tex]First, we solve the additions inside the parenthesis
[tex]36\colon3-2\times5[/tex]Then, we solve the produce and the division
[tex]12-10[/tex]At last, we subtract
[tex]12-10=2[/tex]Hence, the answer is 2.A corporation distributes a 10% common stock dividend on 30000 shares issued when the market value of its common stock is $24 per share and its par value is $2 per share dollars per share on the distribution date a credit for $___ would be journalized.A. $30,000B. $6,000C. $72,000D. $66,000
A corporation distributes a 10% common stock dividend on 30,000 shares.
The market value is $24 per share.
The par value is $2 per share.
We have to find the credit that is journalized the moment the distribution is made.
They paid a total amount in dividends that is 10% of the par value of the stock times the number of stocks:
[tex]\begin{gathered} 10\%\cdot2\cdot30000 \\ 0.1\cdot2\cdot30000 \\ 6000 \end{gathered}[/tex]Answer: the credit is $6,000 [Option B]
this is factor by grouping. did I do 1a right and how do I continue on 1b
We will have the following:
*First: ( f ° g) (x):
[tex](f\circ g)(x)=\frac{(\frac{1}{x})+1}{(\frac{1}{x})-2}\Rightarrow(f\circ g)(x)=\frac{(\frac{1+x}{x})}{(\frac{1-2x}{x})}[/tex][tex]\Rightarrow(f\circ g)(x)=\frac{(1+x)(x)}{(x)(1-2x)}\Rightarrow(f\circ g)(x)=\frac{1+x}{1-2x}[/tex]Domain:
[tex](-\infty,\frac{1}{2})\cup(\frac{1}{2},\infty)[/tex]*Second: (f ° f) (x):
[tex](f\circ f)(x)=\frac{(\frac{x+1}{x-2})+1}{(\frac{x+1}{x-2})-2}\Rightarrow(f\circ f)(x)=\frac{(\frac{(x+1)+(x-2)}{x-2})}{(\frac{(x+1)-2(x-2)}{x-2})}[/tex][tex]\Rightarrow(f\circ f)(x)=\frac{(\frac{2x-1}{x-2})}{(\frac{-x+5}{x-2})}\Rightarrow(f\circ f)(x)=\frac{(2x-1)(x-2)}{(x-2)(-x+5)}[/tex][tex]\Rightarrow(f\circ f)(x)=\frac{2x-1}{-x+5}[/tex]Domain:
[tex](-\infty,5)\cup(5,\infty)[/tex]simplify the following giving the answer with a positive exponent 2n^4*2n^3÷4
so the answer is n^7
please let me know when I come to work with this
Comparing the blue bars (8 - 12 yrs old) and orange bars (13 - 17 yrs old), we can see that most of the blue bars centered between 1 - 1.9 hours of screen time while the orange bars somehow centered between 3 - 3.9 hours of screen time.
If more screen time mean less exercise, then, we can infer that on average, 13 to 17-year-olds gets less exercise compared to 8 to12-year-olds. (Option 3)
On a unit circle, ___ radians. Identify the terminal point andsin f.
Remember the following:
[tex]\begin{gathered} \sin(0)=0 \\ \\ \sin\left(\frac{\pi}{6}\right)=\frac{1}{2} \\ \\ \sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \\ \\ \sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2} \\ \\ \sin\left(\frac{\pi}{2}\right)=1 \end{gathered}[/tex][tex]\begin{gathered} \cos(0)=1 \\ \\ \cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2} \\ \\ \cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \\ \\ \cos\left(\frac{\pi}{3}\right)=\frac{1}{2} \\ \\ \cos\left(\frac{\pi}{2}\right)=0 \end{gathered}[/tex]The terminal point of an angle θ is given by:
[tex](\cos\theta,\sin\theta)[/tex]For θ=π/2, we have:
[tex](\cos\frac{\pi}{2},\sin\frac{\pi}{2})=(0,1)[/tex]Therefore, the answer is: option B) Terminal point: (0,1), sinθ=1.
the function f(x) = |2x-4| is not a one-to-one function. graph the part of the function that is one-to-one and extends to positive infinity.
Here, we want to graph the part of the graph that is one-to-one
What we have to do here is to remove the absolute value signs and plot the graph of the line that it normally looks like
Generally, we have the equation of a straight line as;
[tex]y\text{ = mx + b}[/tex]where m is the slope and b is the y-intercept
Looking at the function f(x) = 2x-4; -4 is simply the y-intercept value
So, we have a point at (0,-4)
To get the second point, set f(x) = 0
[tex]\begin{gathered} 2x-\text{ 4 = 0} \\ 2x\text{ = 4} \\ x\text{ =}\frac{4}{2}\text{ = 2} \end{gathered}[/tex]So, we have the second point as (2,0)
By joining (2,0) to (0,-4) ; we have the plot of the part of the function that extends to infinity
Describe and correct the error in performing the operation of complex numbers and write the answer in standard form.
Answer:
-20+48i
Explanation:
The solution erroneously began by expressing the square as the square of each of the terms.
[tex](4+6i)^2=(4)^2+(6i)^2^{}[/tex]However, the correct way is to take the square of the entire expression inside the bracket as shown below:
[tex](4+6i)^2=(4+6i)(4+6i)[/tex]Next, we expand and simplify our result below:
[tex]\begin{gathered} =4(4+6i)+6i(4+6i) \\ =16+24i+24i+36i^2 \\ =16+48i+36(-1) \\ =16-36+48i \\ =-20+48i \end{gathered}[/tex]The result of the operation in standard form is -20+48i.
what are the roots of the equation?-3= -6x^2+7x
We have the next equation
[tex]-3=-6x^2+7x[/tex]First, we need to set the equation to zero
[tex]6x^2-7x-3=0[/tex]then we will use the general formula to find the roots of a second-degree equation
[tex]x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]where
a=6
b=-7
c=-3
then we substitute the values
[tex]x_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4(6)(-3)}}{2(6)}[/tex][tex]\begin{gathered} x_{1,2}=\frac{7\pm\sqrt[]{49^{}+72}}{12} \\ x_{1,2}=\frac{7\pm\sqrt[]{121}}{12} \\ x_{1,2}=\frac{7\pm11}{12} \\ \end{gathered}[/tex][tex]x_1=\frac{7+11}{12}=\frac{18}{12}=\frac{3}{2}[/tex][tex]x_2=\frac{7-11}{12}=\frac{-4}{12}=-\frac{1}{3}[/tex]the roots of the equation are x=3/2, x=-1/3
A clothing manufacturer has 1,000 yd. of cotton to make shirts and pajamas. A shirt requires 1 yd. of fabric, and a pair of pajamas requires 2 yd. of fabric. It takes 2 hr. to make a shirt and 3 hr. to make the pajamas, and there are 1,600 hr. available to make the clothing. i. What are the variables? ii. What are the constraints? iii. Write inequalities for the constraints. iv. Graph the inequalities and shade the solution set. v. What does the shaded region represent? vi. Suppose the manufacturer makes a profit of $10 on shirts and $18 on pajamas. How would it decide how many of each to make? vii. How many of each should the manufacturer make, assuming it will sell all the shirts and pajamas it makes?
Let the number of shirts is x and the number of pairs of pajamas is y
Then the variables are x and y which are the numbers of shirts and pajamas
Since each shirt needs, 1 yard and a pair of pajamas needs 2 yards
Since there are 1000 yards to make them
Then the first inequality is
[tex]\begin{gathered} (1)x+(2)y\leq1000 \\ x+2y\leq1000 \end{gathered}[/tex]Since the time to make a shirt is 2 hours and the time to make a pair of pajamas is 3 hours
Since there are 1600 hours available, then
The second inequality is
[tex]\begin{gathered} (2)x+(3)y\leq1600 \\ 2x+3y\leq1600 \end{gathered}[/tex]Then let us answer the questions
i. The variables are x and y
ii. The constraints are 1000, 1600
iii. The inequalities are
[tex]\begin{gathered} x+2y\leq1000 \\ 2x+3y\leq1600 \end{gathered}[/tex]iv. Let us draw the graph
The red area represents the 1st inequality
The blue area represents the 2nd inequality
The area of the two colors is the area of the solutions of the 2 inequalities
V.
The shaded region represents the solution of the 2 inequalities, the numbers of shirts and pajams
Vi.
The intersection point between the 2 lines is (200, 400)
Then we will take this point to represents the number of shirts and pajamas
vii.
Since the profit on shirts is $10 and on pajama is $18
Then we should make 200 shirts and 400 pajamas
*Identify the transformations for the function below. Check all that applyf(x) = -3x + 2DilationHorizontal ShiftVertical ShiftAReflection
f (x) = -3x + 2
then
Dilation is 3
Horizontal shift , find 0= -3x +2, x = 2/3
Vertical shift , x= 0 , y=+2
Reflection , find slope m' = -1/m = -1/-3= 1/3
Question 4: -12a - 4 and -4(3a - 1) are equivalent expressions. True False > false
If we use the distributive property on the second expression, we get the following:
[tex]-4\cdot(3a-1)=-4\cdot(3a)-4(-1)=-12a+4[/tex]therefore, the expressions are not equivalent
I need all solved, As soon as possible Question 1
Given:
[tex]f(x)=3^x[/tex]To find:
The type of function by completing the table and graphing the function
Explanation:
When x = -2,
[tex]\begin{gathered} y=3^{-2} \\ =\frac{1}{3^2} \\ =\frac{1}{9} \\ =0.11 \end{gathered}[/tex]When x = -1,
[tex]\begin{gathered} y=3^{-1} \\ =\frac{1}{3} \\ =0.33 \end{gathered}[/tex]When x = 0,
[tex]\begin{gathered} y=3^0 \\ =1 \end{gathered}[/tex]When x = 1,
[tex]\begin{gathered} y=3^1 \\ =3 \end{gathered}[/tex]When x = 2,
[tex]\begin{gathered} y=3^2 \\ =9 \end{gathered}[/tex]Therefore, the table values are,
Then, the graph will be,
Since the domain of the function is real numbers and the range of the function is a set of positive real numbers.
Therefore, it is an exponential function.
Solve equation 1/4 + 1/7=1/t for t to find the number of days it would take them to paint the house if they worked together. Number 361
ANSWER:
2.5 days.
STEP-BY-STEP EXPLANATION:
We have the following equation:
[tex]\frac{1}{4}+\frac{1}{7}=\frac{1}{t}[/tex]We solve for t:
[tex]\begin{gathered} \frac{1\cdot7+4\cdot1}{4\cdot7}=\frac{1}{t} \\ \frac{11}{28}=\frac{1}{t} \\ t=\frac{28}{11}\approx2.5\text{ days} \end{gathered}[/tex]Therefore, if they work together, they could paint the house in about 2.5 days.
8. A farm water tower (with a capacity of 615 cubic metres) has sprung a leak. It loses water at the rate of 1 cubic metre an hour. If no one fixes it, when would the tower be empty? (Answer in weeks, days and hours; for example, 2 weeks, 2 days and 5 hours.)
Given: A farm water tower (with a capacity of 615 cubic metres) has sprung a leak. It loses water at the rate of 1 cubic metre an hour
Find: when would the tower be empty.
Explanation: A capacity of farm water tower is 615 cubic meters.
if it loses water at the rate of 1 cubic meter an hour
it means it take 615 hours to be empty.
[tex]615\text{ hours=}\frac{615}{24}=25.625\text{ days}[/tex]25.625 conatins 3 weeks= 21 days.
25.625-21=4.625 days.
4.625 days contains 4 days and
[tex]0.625\times24=15\text{ hours}[/tex]Hence the final answer will be 3 weeks, 4 days and 15 hours .
QUESTION 6 1 POINTA 20-foot string of lights will be attached to the top of a 12-foot pole for a holiday display. How far from the base of the poleshould the end of the string of lights be anchored?20 AProvide your answer below:ftFEEDBACK+O
EXPLANATION
Since we have the given sides, we can apply the Pythagorean Theorem in order to obtain the needed distance:
[tex]Hypotenuse^2=Larger\text{ side}^2+Smaller\text{ side}^2[/tex]Plugging in the terms into the expression:
[tex]20^2=Larger\text{ side\textasciicircum2+12}^2[/tex]Subtracting 12^2 to both sides:
[tex]20^2-12^2=Larger\text{ side}^2[/tex]Computing the powers:
[tex]400-144=Larger\text{ side}^2[/tex]Subtracting numbers:
[tex]256=Larger\text{ side}^2[/tex]Applying the square root to both sides:
[tex]\sqrt{256}=Larger\text{ side}[/tex]Computing the root:
[tex]16=Larger\text{ side}[/tex]Switching sides:
[tex]Larger\text{ side =16}[/tex]In conclusion, the solution is 16ft
Carbon-14 is used for archeological carbon dating. Its half-life is 5730 years. How much of a 50-gram sample of Carbon-14 will be left in 1000 years?
Given:
The half-life of carbon-14 is 5730 years.
The initial amount of carbon is I = 50 grams.
Explanation:
To find the final amount of carbon after 1000 years.
The fundamental decay equation is,
[tex]\begin{gathered} F=Ie^{-\lambda t} \\ \text{Where, }\lambda=\frac{\ln 2}{t_{\frac{1}{2}}} \end{gathered}[/tex]Let us find the radioactive constant first.
[tex]\begin{gathered} \lambda=\frac{\ln 2}{5730} \\ \lambda=0.00012096809 \end{gathered}[/tex]Then, the final amount of the corban-14 is,
[tex]\begin{gathered} F=50e^{-0.000121(1000)}^{} \\ =44.30g \end{gathered}[/tex]Hence, the amount of a 50-gram sample of Carbon-14 will be left in 1000 years is 44.30 g.
Add.(7g + 4) + (8g + 2)
We have to add the expression.
We will group the similar terms:
[tex]\begin{gathered} \mleft(7g+4\mright)+(8g+2) \\ 7g+8g+4+2 \\ 15g+6 \end{gathered}[/tex]Answer: 15g+6