Answer:
a triangular pyramid
Name some real-life situations where graphing could be useful. Discuss your ideas. Name some real-life situations where finding the coordinates of the midpoint of a line segment could be useful.
Answer:
mapping an area
Step-by-step explanation:
One situation and probably the most common is mapping an area. Graphs are great for dividing a geographical location into various sections and creating a model representation of the area. The graph itself allows for specific directions to be shared using the x and y coordinates on the graph. The same applies for finding the midpoint of a line segment. For example, this is useful if you were trying to find a place to meetup with a friend that is an equal distance from where you are and from where your friend is currently located. Therefore, allowing you to meetup at the midpoint.
Not sure about the answers I gave
Need help with the others
Answer:
Ask me
Step-by-step explanation:
Tell me dear ask..
Given that Q(x)=2x^2 +5x-3 find and simplify Q(a+h)-Q(a-h)
Answer:
[tex]Q(a+h)-Q(a-h)=8ah+10h[/tex]
Step-by-step explanation:
We are given the function:
[tex]Q(x)=2x^2+5x-3[/tex]
And we want to find and simplify:
[tex]Q(a+h)-Q(a-h)[/tex]
Substitute:
[tex]=[2(a+h)^2+5(a+h)-3]-[2(a-h)^2+5(a-h)-3][/tex]
Expand:
[tex]\displaystyle =[2(a^2+2ah+h^2)+5a+5h-3]-[2(a^2-2ah+h^2)+5a-5h-3][/tex]
Distribute:
[tex]=[2a^2+4ah+2h^2+5a+5h-3]-[2a^2-4ah+h^2+5a-5h-3][/tex]
Distribute:
[tex]=(2a^2+4ah+2h^2+5a+5h-3)+(-2a^2+4ah-2h^2-5a+5h+3)[/tex]
Rewrite:
[tex]=(2a^2-2a^2)+(4ah+4ah)+(2h^2-2h^2)+(5a-5a)+(5h+5h)+(-3+3)[/tex]
Combine like terms:
[tex]=8ah+10h[/tex]
Hence:
[tex]Q(a+h)-Q(a-h)=8ah+10h[/tex]
1. You measure 24 textbooks' weights, and find they have a mean weight of 75 ounces. Assume the population standard deviation is 3.3 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.
2. You measure 37 backpacks' weights, and find they have a mean weight of 45 ounces. Assume the population standard deviation is 10.1 ounces. Based on this, construct a 95% confidence interval for the true population mean backpack weight.
3. You measure 30 watermelons' weights, and find they have a mean weight of 37 ounces. Assume the population standard deviation is 4.1 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight.
4. A student was asked to find a 99% confidence interval for widget width using data from a random sample of size n = 16. Which of the following is a correct interpretation of the interval 11.8 < μ < 20.4?
A. There is a 99% chance that the mean of a sample of 16 widgets will be between 11.8 and 20.4.
B. The mean width of all widgets is between 11.8 and 20.4, 99% of the time. We know this is true because the mean of our sample is between 11.8 and 20.4.
C. With 99% confidence, the mean width of all widgets is between 11.8 and 20.4.
D. With 99% confidence, the mean width of a randomly selected widget will be between 11.8 and 20.4.
E. There is a 99% chance that the mean of the population is between 11.8 and 20.4.
5. For a confidence level of 90% with a sample size of 23, find the critical t value.
Answer:
(73.845 ; 76.155) ;
(41.633 ; 48.367) ;
1.273 ;
C. With 99% confidence, the mean width of all widgets is between 11.8 and 20.4. ;
1.717
Step-by-step explanation:
1.)
Given :
Mean, xbar = 75
Sample size, n = 24
Sample standard deviation, s = 3.3
α = 90%
Confidence interval = mean ± margin of error
Margin of Error = Tcritical * s/√n
Tcritical at 90% ; df = 24 - 1 = 23
Tcritical = 1.714
Margin of Error = 1.714 * 3.3/√24 = 1.155
Confidence interval = 75 ± 1.155
Confidence interval = (73.845 ; 76.155)
2.)
Given :
Mean, xbar = 45
Sample size, n = 37
Sample standard deviation, s = 10.1
α = 95%
Confidence interval = mean ± margin of error
Margin of Error = Tcritical * s/√n
Tcritical at 95% ; df = 37 - 1 = 36
Tcritical = 2.028
Margin of Error = 2.028 * 10.1/√37 = 3.367
Confidence interval = 45 ± 3.367
Confidence interval = (41.633 ; 48.367)
3.)
Given :
Mean, xbar = 37
Sample size, n = 30
Sample standard deviation, s = 4.1
α = 90%
Margin of Error = Tcritical * s/√n
Tcritical at 90% ; df = 30 - 1 = 29
Tcritical = 1.700
Margin of Error = 1.700 * 4.1/√30 = 1.273
5.)
Sample size, n = 23
Confidence level, = 90%
df = n - 1 ; 23 - 1 = 22
Tcritical(0.05, 22) = 1.717
How would I simplify the expressions on the picture?
Answer:
7. [tex]x^{11}[/tex] 8. [tex]y^{2}\\[/tex] 9. [tex]p^{12}[/tex] 10.[tex]a^{3} b^{2}[/tex] 11.[tex]g^{16}[/tex] 12.[tex]r^{9} h^{3}[/tex] 13.[tex]m^{15} p^{6}[/tex] 14.[tex]k^{6} y[/tex] 15.[tex]x^6 z^4[/tex]
Step-by-step explanation:
7. [tex]x^3[/tex] × [tex]x^8[/tex] = [tex]x^{11}[/tex] when multiplying with exponents you add
8. [tex]\frac{y^{6} }{y^{4} }[/tex] = [tex]y^{2}[/tex] when dividing with exponents you subtract
9. [tex](p^{3})^4[/tex] = [tex]p^{12}\\[/tex] when it's power to power, you multiply
10. [tex]\frac{a^{9} b^{4}}{a^{6} b^{2}}[/tex] = [tex]a^{3} b^{2}[/tex] (subtract exponents)
11. [tex](g^{8})^2[/tex] = [tex]g^{16}[/tex] (multiply exponents)
12. [tex]r^{4} h^{2} r^{5} h[/tex] = [tex]r^{9} h^{3}[/tex] (add exponents [tex]r^4 + r^5\\[/tex] and [tex]h^2 +h^1\\[/tex] )
13. [tex](m^{5} p^{2})^3[/tex] = [tex]m^{15} p^{6}[/tex] (multiply exponents)
14. [tex]\frac{k^{7} y^{4}}{y^{3}k}[/tex] = [tex]k^{6} y[/tex] (subtract exponents [tex]k^7-k^1[/tex] and [tex]y^4-y^3\\[/tex] )
15. [tex]x^3 z^2 x^3 z^2[/tex] = [tex]x^6 z^4[/tex] (add exponents same as #12)
I will mark you brainliest if you provide evidence you know what your doing
Work out the problem and make the answer clear
Option C
SOLUTION:
We need to find the value of B - CF
First find the value CF:
[tex]CF=\left[\begin{array}{ccc}12&0&1.5\\1&-6&7\\\end{array}\right] \left[\begin{array}{ccc}-2&0\\0&8\\2&1\end{array}\right][/tex]
[tex]CF=\left[\begin{array}{ccc}12(-2)+0 *0+1.5*2&12*0+0.8+1.5*1\\1*(-2)+(-6)*0+7.2&1*0+(-6)*8+7.1\\\end{array}\right][/tex]
[tex]CF=\left[\begin{array}{ccc}-21&1.5\\12&-41\\\end{array}\right][/tex]
Now find value of B - CF:
[tex]B-CF=\left[\begin{array}{ccc}2&8\\6&3\\\end{array}\right] -\left[\begin{array}{ccc}-21&1.5\\12&-41\\\end{array}\right][/tex]
[tex]B-CF=\left[\begin{array}{ccc}23&6.5\\-6&44\\\end{array}\right][/tex]
∴ the value of B - CF is [tex]\left[\begin{array}{ccc}23&6.5\\-6&44\\\end{array}\right][/tex]
I hope this helps....
A town recently dismissed 5 employees in order to meet their new budget reductions. The town had 4 employees over 50 years of age and 16 under 50. If the dismissed employees were selected at random, what is the probability that no more than 1 employee was over 50
Answer:
0.7513 = 75.13% probability that no more than 1 employee was over 50
Step-by-step explanation:
The employees are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
The probability of x successes is given by the following formula:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
In which:
x is the number of successes.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this question:
4 + 16 = 20 employees, which means that [tex]N = 20[/tex]
4 over 50, which means that [tex]k = 4[/tex]
5 were dismissed, which means that [tex]n = 5[/tex]
What is the probability that no more than 1 employee was over 50?
Probability of at most one over 50, which is:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
In which
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 0) = h(0,20,5,4) = \frac{C_{4,0}*C_{16,5}}{C_{20,5}} = 0.2817[/tex]
[tex]P(X = 1) = h(1,20,5,4) = \frac{C_{4,1}*C_{16,4}}{C_{20,5}} = 0.4696[/tex]
Then
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.2817 + 0.4696 = 0.7513[/tex]
0.7513 = 75.13% probability that no more than 1 employee was over 50
Square Footage Frequency
0-499 5
500-999 17
1000-1499 36
1500-1999 115
2000-2499 125
2500-2999 81
3000-3499 47
3500-3999 45
4000-4499 22
4500-4999 7
The accompanying frequency distribution represents the square footage of a random sample of 500 houses that are owner occupied year round. Approximate the mean and standard deviation square footage.
Answer:
2424.5
904.16
Step-by-step explanation:
the mean = ∑frequency /n
∑f = 5+17+36+115+125+81+47+45+22+7 = 500
∑xf = 1212250
∑x²f = 3347037625
sample mean = 1212250/500
= 2424.5
variance = 1/500-1[3347037625 - 1212250²]
= 815710.02
standard deviation is = √variance
standard deviation = √815710.02
= 904.16
Find the perimeter of the
polygon if ZB = D.
3 om
B
4 cm
D
5 cm
C
P = [?] cm
Answer:
16 cm
Step-by-step explanation:
4 + 4 + 3 + 5 = 16
The = sign means that B (which is 4 cm) is equal to D (which had no number)
And because it says that B = D (with the squiggly line (or a tilde)) And the L's (which means that the letters represent an angle) All you have to do is add the numbers together, and you get 16.
Sorry if I explained it badly, you at least got the answer.
(And also, if I'm wrong, please tell me.)
Answer:
P = 32 cm
Step-by-step explanation:
Im just putting the right answer up so you don't accidentally put in the wrong one.
Which division problem does the diagram below best illustrate?
A diagram with 8 ovals containing 4 squares each.
O 16 divided by 4 = 4
O 32 divided by 4 = 8
O 36 divided by 4 = 9
O 8 divided by 2 = 4
Answer:
The answer is 32 divided by 4
Step-by-step explanation:
Because in each box there is 4. There are 8 ovals all together. So 8×4, you get 32 and divide it by the number of squares in an oval which is 4
Answer:
the answer is 32 divided by 4=8
Step-by-step explanation:
because when you look at the ovals there's eight ovals and in side there's four squares..
HOPE THIS HELPS!!!!!
Which of the following is equivalent to the expression log2a=r? 2a = r logr2 = a 2r = a log2r = a
9514 1404 393
Answer:
(c) 2^r = a
Step-by-step explanation:
The relationship between log forms and exponential forms is ...
[tex]\log_2(a)=r\ \Leftrightarrow\ 2^r=a[/tex]
__
Additional comment
I find this easier to remember if I think of a logarithm as being an exponent.
Here, the log is r, so that is the exponent of the base, 2.
This equivalence can also help you remember that the rules of logarithms are very similar to the rules of exponents.
Answer: Choice C) [tex]2^r = a[/tex]
This is the same as writing 2^r = a
==========================================================
Explanation:
Assuming that '2' is the base of the log, then we'd go from [tex]\log_2(a) = r[/tex] to [tex]2^r = a[/tex]
In either equation, the 2 is a base of some kind. It's the base of the log and it's the base of the exponent.
The purpose of logs is to invert exponential operations and help isolate the exponent. A useful phrase to help remember this may be: "if the exponent is in the trees, then we need to log it down".
The general rule is that [tex]\log_b(y) = x[/tex] converts to [tex]y = b^x[/tex] and vice versa.
Can you please help me with this question
Joe bikes at the speed of 30 km/h from his home toward his work. If Joe's wife leaves home 5 mins later by car, how fast should she drive in order to overtake him in 10 minutes.
Answer:
45 mph
Step-by-step explanation:
This is a really good question to know the answer to. It is tricky and a bit indirect (which means you have to find something else before you can find the speed of the car.)
Let's keep track of what he does in the time allotted.
How far does Joe go in 5 minutes? That's the amount of time he's on the road before she is.
convert 5 minutes into hours. 5 minutes * 1 hour / 60 minutes = 1/12 of an hour
d = r*t
r = 30 km/hour
t = 1/12 hour
d = 30 km/hr * 1/12 hour = 2.5 km
Now she's about to start. She wants to catch him in 10 minutes
d = r*t
r = x mph
t = 10 minutes = 10 minutes * 1 hour * 60 minutes = 1/6 of an hour.
How far does he go in the 10 minute time?
d = 30 * 1/6 = 5 km
What is his total distance
5 km + 2.5 km = 7.5 km
Finally how fast does she need to go to catch him
d = 7.5 km
r = ? This is what you are trying to find
t = 1/6 of an hour
d = r*t
7.5 km = r * (1/6)hour divide by 1/6 hour
7.5 km // 1/6 hour = r
r = 7.5 * 6 = 45 mph
I need help solving this problem .
Step-by-step explanation:
here is the answer to your question
If you select a random sample of 6 people, what is the probability that all but one of them are using Chrome as their browser? (show all work - use data below)
Use the data provided below:
Chrome: 52.57%
Safari: 31.17%
Firefox: 4.29%
Edge: 2.91%
IE: 2.47%
Samsung: 2.36%
Answer:
0.1143 = 11.43% probability that all but one of them are using Chrome as their browser
Step-by-step explanation:
For each person, there are only two possible outcomes. Either they use Chrome, or they do not. The probability of a person using Chrome is independent of any other person, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Chrome: 52.57%
This means that [tex]p = 0.5257[/tex]
Sample of 6 people
This means that [tex]n = 6[/tex]
What is the probability that all but one of them are using Chrome as their browser?
5 using Chrome, so:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{6,5}.(0.5257)^{5}.(1-0.5257)^{1} = 0.1143[/tex]
0.1143 = 11.43% probability that all but one of them are using Chrome as their browser
In the accompanying diagram of isosceles triangle ABC, overline AB cong overline BC , BAC =X , and m angle ABC=3x+70
Answer:
x = 22
Step-by-step explanation:
In order to solve this, we need to understand that in an isosceles triangle the two angles that are located at its base are equal to each other.
base - (the side that is not one of the two sides that are equivalent to each other)
Knowing this we can see that ∠ACB will equal ∠BAC, therefore ∠ACB will be equal to x°. Since the sum of all inner angles of a triangle is equal to 180°, we can make the following equation...
x° + x° + (3x + 70)° = 180°
2x° + 3x° + 70° = 180°
5x° = 180° - 70°
5x° = 110°
x° = 110° / 5
x° = 22°
x = 22
Therefore, x = 22.
A loan of £1000 has a compound interest rate of 2.7% charged monthly. Express the original loan as a percentage of the total amount awed after 2 months if no payment are made
Answer:
£1054.729
Step-by-step explanation:
To find compound interest you need to use the equation 1000(1.027)^x.
To find the interest rate (1.027):
100 + 2.7 = 102.7
102.7 / 100 = 1.027
The value of x is the amount of months if no payment is made in this situation, so 2 would be the x value for this problem.
Hope this helps!
A large container holds 4 gallons of chocolate milk that has to be poured into bottles. Each bottle holds 2 pints.
If the ratio of gallons to pints is 1: 8,
bottles are required to hold the 4 gallons of milk.
Answer:
64 Bottles
Step-by-step explanation:
that is the procedure above
A point is selected at random from a line segment of length l, dividing it into two line segments. What is the probability that the longer line segment is at least three times as long as the shorter segment
Answer:
3/4
Step-by-step explanation:
Let a be the length of the shorter line segment and b be the length of the longer line segment.
Since the length of the line segment is l, we have that the length of the line segment equals length of shorter line segment + length of longer line segment.
So, l = a + b
Since we require that the longer line segment be at least three times longer than the shorter line segment, we have that b = 3a
So, l = a + b
l = a + 3a
l = 4a
The probability that the shorter line segment will be a(or 3 times shorter than b) is P(a) = length of shorter line segment/length of line segment = a/l
Since l = 4a.
a/l = 1/4
So, P(a) = 1/4
The probability that a will be less than 3 times shorter that b is P(a ≤ 1) = P(0) + P(a) = 0 + 1/4 = 1/4
The probability that b will be 3 times or more greater than a is thus P(b ≥ 3) = 1 - P(a ≤ 1) = 1 - 1/4 = 3/4
would someone mind looking over my answers to geometry!!
Answer:
Question 1: x = 6
Question 2: Correct!
Question 3: x = 11
Question 4: Correct!
Step-by-step explanation:
Question 1:
Angle 22x - 2 DOESN'T equal 50 degrees. Only Alternate Interior Angles will equal each other. These two angles are Same Side Interior Angles, meaning if you added them together, they would equal 180 degrees.
Knowing that adding 22x - 2 and 50 will equals 180 degrees, here's how we solve for x:
First, subtract 50 from 180 to find what angle 22x - 2 will equal:
180 - 50 = 130
130 = 22x - 2
Now use basic algebra to solve for x:
130 = 22x - 2
(add 2 to both sides)
132 = 22x
(divide both sides by 22)
x = 6
Question 3:
Angle 5x + 15 DOESN'T equal 9x + 11. They make up a line, which is 180 degrees, so they are supplementary angles.
With that in mind, to solve for x, add the two equations and set it equal to 180:
5x + 15 + (9x + 11) = 180
Now use basic algebra to solve for x:
5x + 15 + 9x + 11 = 180
(add like terms)
14x + 26 = 180
(subtract 26 from both sides)
14x = 154
(divide 14 from both sides)
x = 11
Hope it helps (●'◡'●)
4
5
start fraction, 5, divided by, 4, end fraction hour ==equals
minutes
Answer:
1.25. It would be 1.25 if ur just talking about dividing in general which is pretty tough
Answer:
\dfrac54=-4c+\dfrac14 4 5 =−4c+ 4 1 start fraction, 5, divided by, 4, end fraction, equals, minus, 4, c, plus, start fraction, 1, divided by, 4, end fraction
Step-by-step explanation:
In one state lottery game, you must select four digits (digits may be repeated). If your number matches exactly the four digits selected by the lottery commission, you win.
1) How many different numbers may be chosen?
2) If you purchase one lottery ticket, what is your chance of winning?
3) There are ___ different numbers that can be chosen. (Type a whole number.)
4) There is a ___ chance of winning.*
*The answer choices for number 4 are:
1 in 10,000
1 in 6,561
1 in 100
1 in 1,000
1 in 9,999
Answer:
Part 1)
10,000 different numbers.
Part 2)
A) 1 in 10,000.
Step-by-step explanation:
Part 1)
Since there are four digits and there are ten choices for each digit (0 - 9) and digits can be repeated, then we will have:
[tex]T=\underbrace{10}_{\text{Choices For First Digit}}\times\underbrace{10}_{\text{Second Digit}}\times\underbrace{10}_{\text{Third Digit}}\times \underbrace{10}_{\text{Fourth Digit}} = 10^4=10000[/tex]
Thus, 10,000 different numbers are possible.
Part 2)
Since there 10,000 different tickets possible, the chance of one being the correct combination will be 1 in 10,000.
This is equivalent to 0.0001 or a 0.01% chance of winning.
(-3).(+9)-(-24)-(+6).(+2)
9. Mariah has 28 centimeters of reed
and 37/100 meters of reed for weaving
baskets. How many meters of reed
does she have? Write your answer as a
decimal and explain your answer. (The first time time I asked I forgot to put the 37/100)
Answer:
0.65m
Step-by-step explanation:
28cm is equal to 0.28m
37/100 is 37% of a metre so 0.37m
0.28 + 0.37 = 0.65m
Levi makes the minimum salary for actuary. Andres maybe the median salary for cpa. Who makes more money
Answer:
Andres
why?
Because he is median salary for cpa
A radioactive substance decays exponentially: The mass at time t is m(t) = m(0)e^kt, where m(0) is the initial mass and k is a negative constant. The mean life M of an atom in the substance is
[infinity]
M = âk â« te^kt dt.
0
For the radioactive carbon isotope, 14C, used in radiocarbon dating, the value of k is -0.000121. Find the mean life of a 14C atom.
Answer:
mean life = 8264.5 s
Step-by-step explanation:
k = - 0.000121
The relation is given by
[tex]m = mo e^{kt}[/tex]
Now, the mean life is the life time for which the sample retains.
The mean life is the reciprocal of the decay constant.
The relation between the mean life and the decay constant is
[tex]\tau =\frac{1}{k}\\\\\tau = \frac{1}{0.000121} = 8264.5 seconds[/tex]
In a pool of water filled to a depth of 10 ft, calculate the fluid force on one side of a 3 ft by 4 ft rectangular plate if it rests vertically on its 4 ft edge at the bottom of the pool. Remember that water weighs 62.4 lb/ft3
9514 1404 393
Answer:
6,364.8 lb
Step-by-step explanation:
The centroid of the plate is its center, so is 1.5 ft above the bottom of the pool, or 8.5 ft below the surface. The area of the plate is (3 ft)(4 ft) = 12 ft². Then the fluid force is ...
(62.4 lb/ft³)(8.5 ft)(12 ft²) = 6,364.8 lb
a + b·c = a + c·b is an example of the associative property.
Answer:
yes , This is an example of the associative property.
Step-by-step explanation:
A cell site is a site where electronic communications equipment is placed in a cellular network for the use of mobile phones. The numbers y of cell sites from 1985 through 2011 can be modeled byy = 269573/1+985e^-0.308t where t represents the year, with t = 5 corresponding to 1985. Use the model to find the numbers of cell sites in the years 1998, 2003, and 2006.
Answer:
(a) 3178
(b) 14231
(c) 33152
Step-by-step explanation:
Given
[tex]y = \frac{269573}{1+985e^{-0.308t}}[/tex]
Solving (a): Year = 1998
1998 means t = 8 i.e. 1998 - 1990
So:
[tex]y = \frac{269573}{1+985e^{-0.308*8}}[/tex]
[tex]y = \frac{269573}{1+985e^{-2.464}}[/tex]
[tex]y = \frac{269573}{1+985*0.08509}[/tex]
[tex]y = \frac{269573}{84.81365}[/tex]
[tex]y = 3178[/tex] --- approximated
Solving (b): Year = 2003
2003 means t = 13 i.e. 2003 - 1990
So:
[tex]y = \frac{269573}{1+985e^{-0.308*13}}[/tex]
[tex]y = \frac{269573}{1+985e^{-4.004}}[/tex]
[tex]y = \frac{269573}{1+985*0.01824}[/tex]
[tex]y = \frac{269573}{18.9664}[/tex]
[tex]y = 14213[/tex] --- approximated
Solving (c): Year = 2006
2006 means t = 16 i.e. 2006 - 1990
So:
[tex]y = \frac{269573}{1+985e^{-0.308*16}}[/tex]
[tex]y = \frac{269573}{1+985e^{-4.928}}[/tex]
[tex]y = \frac{269573}{1+985*0.00724}[/tex]
[tex]y = \frac{269573}{8.1314}[/tex]
[tex]y = 33152[/tex] --- approximated
A bricklayer needs to order 6 300 kg of building sand.
a) Write 6 300 kg in grams, giving your answer in standard form.
One grain of this sand approximately weighs 7 x 10°g.
b) How many grains of sand are there in 6 300 kg of sand? Give your answer in standard from.
Answer:
It would be 6300000. I can't write this in standard form.
Step-by-step explanation:
Answer:
6.3 x 10^6
Step-by-step explanation: