Answer:
Tom:
initial money = $ 39000
% increased per annum = 2.9%
money gained per annum = 39000 * 2.9/100 = $1131
Allen:
initial money = $ 39000
% increased per annum = 5.7 %
money gained per annum = 39000 * 5.7/100 = $2223
Allen has $ (2223 - 1131) = $ 1192 more than Tom
which of the following not between -10 and -8
-17/2
-7
-9
-8.5
Answer:
-7Step-by-step explanation:
This is best read on the number line.
Look at the picture.
[tex]-\dfrac{17}{2}=-8\dfrac{1}{2}=-8.5[/tex]
one third multiplied by the sum of a and b
Answer:
1/3(a+b)
hope it helps :>
Find a cubic polynomial with integer coefficients that has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.
Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]
$a^{2}=5+2 \sqrt{6}$
$a^{3}=11 \sqrt{2}+9 \sqrt{3}$
The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.
Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$
so fits with the other answers.
Answer:
[tex]y^3 -6y-6[/tex]
Find the principal invested if $495 interest was earned in 3 years at an interest rate of 6%.
Answer: $2750
Step-by-step explanation:
Formula to calculate interest : I = Prt , where P = Principal amount , r = rate of interest ( in decimal) , t= time.
Given: I= $495
t= 3 years
r= 6% = 0.06
Then, according to the above formula:
[tex]495 = P (0.06\times3)\\\\\Rightarrow\ P=\dfrac{495}{0.18}\\\\\Rightarrow\ P=2750[/tex]
Hence, the principal invested = $2750
A sprinkler system is being installed in a newly renovated building on campus. The average activation time is supposed to be at most 20 seconds. A series of 12 fire alarm/sprinkler system tests results in an average activation time of 21.5 seconds. Do these data indicate that the design specifications have not been met? The hypotheses to be tested are H0: m = 20 versus Ha: m > 20, where m = the true average activation time of the sprinkler system. Assume that activation times for this system are Normally distributed with s = 3 seconds.
(a) What is the value of the observed test statistic?
(b) What is the value of the P-value?
(c) Are the data statistically significant at the 5% significance level? Explain briefly.
(d) What does the decision you made mean with respect to the question "Do these data indicate that the design specifications have not been met?"
(e) If the true average activation time of the sprinkler system is, in fact, equal to 20 seconds, what type of error would you have made?
Answer:
A) t = 1.73
B) p-value = 0.0558
C) Data is not statistically significant because the p-value of 0.0558 is more than the significance value of 0.05
D) The decision means that the design specifications are not met.
E) Type II error
Step-by-step explanation:
The hypotheses are:
H₀: μ = 20
H₁: μ > 20
A) Formula for the test statistic is;
t = (x' - μ)/(s/√n)
Now, we are given;
x' = 21.5
μ = 20
s = 3
n = 12
Thus;
t = (21.5 - 20)/(3/√12)
t = 1.73
B) we have our t-value as 1.73
Now, Degree of freedom(DF) = n - 1
So,DF = 12 - 1 = 11
Using significance level of α = 0.05, t-value = 1.73 and DF = 11, one tailed hypothesis, from online P-value calculator attached, we have;
p-value = 0.0558
C) Data is not statistically significant because the p-value of 0.0558 is more than the significance value of 0.05
D) We will not reject the null hypothesis. The decision means that the design specifications are not met.
E) If the true average activation time of the sprinkler system is, in fact, equal to 20 seconds, then the null hypothesis is false.
Since we did not reject the null hypothesis even though it is false, the error that was committed was therefore a type II error.
Hey market sales six cans of food for every seven boxes of food the market sold a total of 26 cans and boxes today how many of each kind did the market sale
Answer:
It sold 14 cans boxes of food and 12 cans of food.
Step-by-step explanation:
The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.
Let the no. of sets of food boxes be x.
According to the question,
6x+7x=26
13x=26
x=26/13
x=2
No. of food cans =6x=6×2=12 cans
No. of food boxes=7x=7×2=14 boxes
Please mark brainliest ,if it is truly the best ! Thank you!
logx-log(x-l)^2=2log(x-1)
Answer:
x = 1.00995066776
x = 2.52925492433
Step-by-step explanation:
This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...
[tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]
The attached graph shows zeros at
x = 1.00995066776 and 2.52925492433
_____
Comment on the equation
Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068
Comment on the answer refinement
We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.
In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal places.)
a. x=5
b. x <= 5
c. x>=6
Answer:
[tex]\mathbf{P(X=5) =0.0888}[/tex]
P(x ≤ 5 ) = 0.9707
P ( x ≥ 6) = 0.0293
Step-by-step explanation:
The probability of a binomial mass distribution can be expressed with the formula:
[tex]\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}[/tex]
[tex]\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}[/tex]
where;
n = 8 and π = 0.36
For x = 5
The probability [tex]\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}[/tex]
[tex]\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}[/tex]
[tex]\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}[/tex]
[tex]\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}[/tex]
[tex]\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}[/tex]
[tex]\mathtt{P(X=5) =0.0887645}[/tex]
[tex]\mathbf{P(X=5) =0.0888}[/tex] to 4 decimal places
b. x ≤ 5
The probability of P ( x ≤ 5)[tex]\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})[/tex]
[tex]{P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +[/tex][tex]\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )[/tex]
P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888
P(x ≤ 5 ) = 0.9707
c. x ≥ 6
The probability of P ( x ≥ 6) = 1 - P( x ≤ 5 )
P ( x ≥ 6) = 1 - 0.9707
P ( x ≥ 6) = 0.0293
Determine the value of x in the figure. Question 1 options: A) x = 90 B) x = 85 C) x = 45 D) x = 135
Answer:
A.) x=90°
Step-by-step explanation:
Note:
The triangle shown is an isosceles triangle, which means that it has 2 congruent sides (as shown by the small intersecting lines), and this also means that it has two congruent angles.
We are given an angle measure adjacent to one of the missing angles. These two form supplementary angles, which means that they're sum is equal to 180°, or a straight line. So, to find:
[tex]180=135+y[/tex]
y is the unknown angle. Solve for y:
[tex]180-135=y\\\\y=45[/tex]
y is 45°. Since this and the other angle are congruent, add:
[tex]45+45=90[/tex]
Note:
Triangles angles will always add up to a total of 180°.
To find the missing angle x°, use:
[tex]180=a+b+c[/tex]
These are the angles in a triangle. Substitute any known values and solve:
[tex]180=45+45+x\\\\180=90+x\\\\180-90=x\\\\x=90[/tex]
The missing angle x° is 90°.
:Done
PLEASE HURRY! i walked north 8 miles, the west 4 miles, and finally south 5 miles, at the end how far was i from where i started
Answer:
5 miles away
Step-by-step explanation:
If you walked north 8 miles, then west 4 miles, then south 5 miles, you have, in total, travelled 4 miles west and [tex]8-5=3[/tex] miles north.
This creates a triangle, in which we can find the the length of the hypotenuse to find how far away you are now.
We can use the Pythagorean theorem since this is a right triangle.
[tex]a^2+b^2=c^2\\3^2+4^2=c^2\\9+16=c^2\\25=c^2\\c=5[/tex]
Hope this helped!
Answer:
5 miles away
Step-by-step explanation:
cooks are needed to prepare for a large party. Each cook can bake either 5 Large cakes or 14 small cakes per hour . The kitchen is available for 3 hours and 29 large cakes and 260 cakes need to be baked . How many cooks are required to bake the required number of cakes during the time the kitchen is available?
it was all about equating some values
to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.
Let's determine the number of cooks required to bake the required number of cakes during the available time.
We have the following information:
- Each cook can bake either 5 large cakes or 14 small cakes per hour.
- The kitchen is available for 3 hours.
- We need to bake 29 large cakes and 260 cakes in total.
First, let's calculate the number of large cakes that can be baked by one cook in 3 hours:
1 cook can bake 5 large cakes/hour × 3 hours = 15 large cakes.
Next, let's calculate the number of small cakes that can be baked by one cook in 3 hours:
1 cook can bake 14 small cakes/hour × 3 hours = 42 small cakes.
Now, let's calculate the number of large cakes that can be baked by all the cooks in 3 hours:
Total number of large cakes = Number of cooks × Large cakes per cook per 3 hours
We need to bake 29 large cakes, so:
29 = Number of cooks × 15
Number of cooks = 29 / 15 ≈ 1.93
Since we can't have a fraction of a cook, we need to round up to the nearest whole number. Therefore, we need at least 2 cooks to bake the required number of large cakes.
Similarly, let's calculate the number of small cakes that can be baked by all the cooks in 3 hours:
Total number of small cakes = Number of cooks × Small cakes per cook per 3 hours
We need to bake 260 small cakes, so:
260 = Number of cooks × 42
Number of cooks = 260 / 42 ≈ 6.19
Again, rounding up to the nearest whole number, we need at least 7 cooks to bake the required number of small cakes.
Since we need to satisfy both requirements for large and small cakes, we choose the larger number of cooks required, which is 7 cooks.
Therefore, to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.
Learn more about work here
https://brainly.com/question/13245573
#SPJ2
Use Lagrange multipliers to find three numbers whose sum is 30 and the product P = x3y4z is a maximum. Choose the answer for the smallest of the three values. Question 20 options: a) 21/4 b) 5 c) 15/4 d) 3
We want to maximize [tex]x^3y^4z[/tex] subject to the constraint [tex]x+y+z=30[/tex].
The Lagrangian is
[tex]L(x,y,z,\lambda)=x^3y^4z-\lambda(x+y+z-30)[/tex]
with critical points where the derivatives vanish:
[tex]L_x=3x^2y^4z-\lambda=0[/tex]
[tex]L_y=4x^3y^3z-\lambda=0[/tex]
[tex]L_z=x^3y^4-\lambda=0[/tex]
[tex]L_\lambda=x+y+z-30=0[/tex]
[tex]\implies\lambda=3x^2y^4z=4x^3y^3z=x^3y^4[/tex]
We have
[tex]3x^2y^4z-4x^3y^3z=x^2y^3z(3y-4x)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}\\z=0,\text{ or}\\3y=4x\end{cases}[/tex]
[tex]3x^2y^4z-x^3y^4=x^2y^4(3z-x)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}\\3z=x\end{cases}[/tex]
[tex]4x^3y^3z-x^3y^4=x^3y^3(4z-y)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}4z=y\end{cases}[/tex]
Let's work with [tex]x=3z[/tex] and [tex]y=4z[/tex], for which we have
[tex]x+y+z=8z=30\implies z=\dfrac{15}4\implies\begin{cases}x=\frac{45}4\\y=15\end{cases}[/tex]
The smallest of these is C. 15/4.
Suppose that a sample mean is .29 with a lower bound of a confidence interval of .24. What is the upper bound of the confidence interval?
Answer:
The upper bound of the confidence interval is 0.34
Step-by-step explanation:
Here in this question, we want to calculate the upper bound of the confidence interval.
We start by calculating the margin of error.
Mathematically, the margin of error = 0.29 -0.24 = 0.05
So to get the upper bound of the confidence interval, we simply add this margin of error to the mean
That would be 0.05 + 0.29 = 0.34
Type the missing number in this sequence:
1,
4,
,64, 256,
1,024
Answer:
16
Step-by-step explanation:
The sequence is 1, 4,...,64, 256, 1024
Notice that:
● 1 = 2^0
● 4 = 2^2
● 64 = 2^6
● 256 = 2^8
● 1024 = 2^10
Notice that we add 2 each time to the exponent so the missing number is:
● 2^(2+2) = 2^4 = 16
The age of some lecturers are 42,54,50,54,50,42,46,46,48 and 48 calculate the mean age and standard deviation
Answer:
Mean age: 48
Standard deviation: 4
Step-by-step explanation:
a) Mean
The formula for Mean = Sum of terms/ Number of terms
Number of terms
= 42 + 54 + 50 + 54 + 50 + 42 + 46 + 46 + 48+ 48/ 10
= 480/10
= 48
The mean age is 48
b) Standard deviation
The formula for Standard deviation =
√(x - Mean)²/n
Where n = number of terms
Standard deviation =
√[(42 - 48)² + (54 - 48)² + (50 - 48)² +(54 - 48)² + (50 - 48)² +(42 - 48)² + (46 - 48)² + (46 - 48)² + (48 - 48)² + (48 - 48)² / 10]
= √-6² + 6² + 2² + 6² + 2² + -6² + -2² + -2² + 0² + 0²/10
=√36 + 36 + 4 + 36 + 4 + 36 + 4 + 4 + 0 + 0/ 10
=√160/10
= √16
= 4
The standard deviation of the ages is 4
You are investing $5,000 and can invest for 2 years or 3 years at 1.75% and 1.25% interest rates, respectively. Which earns more interest
Answer:
The 3 years investment earns more interest
Step-by-step explanation:
Given
Principal, P = $5,000
Required
Determine which earns more interest
When Rate = 1.75% and Year = 2
Interest is as follows;
[tex]I = \frac{PRT}{100}[/tex]
Substitute 1.75 for R, 5000 for P and 2 for T
[tex]I = \frac{5000 * 1.75 * 2}{100}[/tex]
[tex]I = \frac{17500}{100}[/tex]
[tex]I = \$175[/tex]
When Rate = 1.25% and Year = 3
Interest is as follows;
[tex]I = \frac{PRT}{100}[/tex]
Substitute 1.25 for R, 5000 for P and 3 for T
[tex]I = \frac{5000 * 1.25 * 3}{100}[/tex]
[tex]I = \frac{18750}{100}[/tex]
[tex]I = \$187.5[/tex]
Comparing the interest of both investments, the 3 years investment earns more interest
solve this equation 4log√x - log 3x =log x^2
Answer:
[tex]x = \frac{1}{3} [/tex]
Step-by-step explanation:
*Move terms to the left and set equal to zero:
4㏒(√x) - ㏒(3x) - ㏒(x²) = 0
*simplify each term:
㏒(x²) - ㏒(3x) - ㏒(x²)
㏒(x²÷x²) -㏒(3x)
㏒(x²÷x² / 3x)
*cancel common factor x²:
㏒([tex]\frac{1}{3x}[/tex])
*rewrite to solve for x :
10⁰ = [tex]\frac{1}{3x}[/tex]
1 = [tex]\frac{1}{3x}[/tex]
1 · x = [tex]\frac{1}{3x}[/tex] · x
1x = [tex]\frac{1}{3}[/tex]
*that would be our answer, however, the convention is to exclude the "1" in front of variables so we are left with:
x = [tex]\frac{1}{3}[/tex]
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of ounces and a standard deviation of ounce. You randomly select cans and carefully measure the contents. The sample mean of the cans is ounces. Does the machine need to be reset? Explain your reasoning. ▼ Yes No , it is ▼ very unlikely likely that you would have randomly sampled cans with a mean equal to ounces, because it ▼ lies does not lie within the range of a usual event, namely within ▼ 1 standard deviation 2 standard deviations 3 standard deviations of the mean of the sample means.
Complete question is;
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.
(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.
Answer:
Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.
Step-by-step explanation:
We are given;
Mean: μ = 128
Standard deviation; σ = 0.2
n = 35
Now, formula for standard error of mean is given as;
se = σ/√n
se = 0.2/√35
se = 0.0338
Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;
μ ± 2se = 128 ± 0.0338
This gives; 127.9662, 128.0338
So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.
Consider the function below. (If an answer does not exist, enter DNE.) f(x) = x3 − 27x + 3 (a) Find the interval of increase. (Enter your answer using interval notation.)
Answer:
(-∞,-3) and (3,∞)
Step-by-step explanation:
f(x) = x³ − 27x + 3
1. Find the critical points
(a) Calculate the first derivative of the function.
f'(x) = 3x² -27
(b) Factor the first derivative
f'(x)= 3(x² - 9) = 3(x + 3) (x - 3)
(c) Find the zeros
3(x + 3) (x - 3) = 0
x + 3 = 0 x - 3 = 0
x = -3 x = 3
The critical points are at x = -3 and x = 3.
2. Find the local extrema
(a) x = -3
f(x) = x³ − 27x + 3 = (-3)³ - 27(-3) + 3 = -27 +81 + 3 = 57
(b) x = 3
f(x) = x³ − 27x + 3 = 3³ - 27(3) + 3 = 27 - 81 + 3 = -51
The local extrema are at (-3,57) and (3,-51).
3, Identify the local extrema as maxima or minima
Test the first derivative (the slope) over the intervals (-∞, -3), (-3,3), (3,∞)
f'(-4) = 3x² -27 = 3(4)² - 27 = 21
f'(0) = 3(0)² -27 = -27
f'(4) = 3(4)² - 27 = 51
The function is increasing on the intervals (-∞,-3) and (3,∞).
The graph below shows the critical points of your function.
At a high school movie night, the refreshments stand sells popcorn and soft drinks. Of the 100 students who came to the movie, 62 bought popcorn and 47 bought a drink. 38 students bought both popcorn and a drink. What is the probability that a student buys a drink, given that he or she buys popcorn? Express your answer as a percent, rounded to the nearest tenth... best answer wins brainliest!!!
Answer:
47% and 62%
Step-by-step explanation:
1. Drink
The probability that a student buys a drink is 0.47
Step-by-step explanation:
The probability that a student buys a drink will be given by;
( the number of students who bought a drink)/(the total number of students)
We are told that;
Of the 100 students who came to the movie, 62 bought popcorn and 47 bought a drink. Therefore, the required probability is;
47/100= 0.47
0.47 = 47%
2. Popcorn
For popcorn probability, it's basically the same.
The probabilty that a student buys popcorn is 0.62
The probability that a student buys popcorn will be given by;
( the number of students who bought popcorn)/(the total number of students)
So therefore,
62/100 = 0.62
0.62 = 62%
What is the mulitplicative rate of change for the exponential function f(x) = 2 (5over2) to the negative x power ?
Answer:
2/5
Step-by-step explanation:
f(x) = 2(5/2)^-x = 2(2/5)^x
The multiplicative rate of change is the base of the positive exponent, 2/5.
Prove that if a and b are integers, then for any integer k one has (a,b) = (a + kb,b). (Hint: Show that they are mutually divisible.)
Answer:
The operation:
(a,b) is equal to the rest of the division of a by b.
Now, if we have:
(a + kb,b) = (a,b) + (k*b,b)
But if we have that k and b are integers, then:
(k*b)/b = k
So b divides k*b into a whole number, this means that (k*b,b) = 0
then:
(a + kb,b) = (a,b) + (k*b,b) = (a,b) + 0 = (a,b)
At a local high school, the student population is growing at 12% a year. If the original population was 242 students, how long will it take the population to reach 300 students? Round to the nearest tenth of a year.
Answer: 2 years
Step-by-step explanation:
The exponential growth function is given by :-
[tex]y=A(1+r)^x[/tex] (i)
, where A = initial value , r = rate of growth and x= time period.
As per given ,
A= 242
r= 12% = 0.12
To find : t when y= 300.
Put all the values in (i)
[tex]300=242(1+0.12)^x\\\\\Rightarrow\ \dfrac{300}{242}=(1.12)^x\\\\\Rightarrow\ 1.23967=(1.12)^x[/tex]
Taking log on both sides , we get
[tex]\log (1.2396) = t \log (1.12)\\\\\Rightarrow\ 0.09328=t(0.049218)\\\\\Rightarrow t=\dfrac{0.09328}{0.049218}=\approx2[/tex]
hence, it will take 2 years.
A bag of 100 hard candies included 30 butterscotch, 40 peppermint, 15 strawberry, 10 orange, and 5 banana. The probability that the first candy pulled out of the bag will be butterscotch or strawberry is .45
a) true
b) false
Answer:
true
Step-by-step explanation:
there is 100 candies. That means we can easily turn the amount of each type of candy into a percent. there was 30 butterscotch which means that is 30 percent. There was 15 strawberry which means that is 15 percent. add that and you get 45. This is a shortcut and i advise you use the way your teacher taught you.
[tex]|\Omega|=100\\|A|=30+15=45\\\\P(A)=\dfrac{45}{100}=0.45[/tex]
So TRUE
You are ordering two pizzas. A pizza can be small, medium, large, or extra large, with any combination of 8 possible toppings (getting no toppings is allowed, as is getting all 8). How many possibilities are there for your two pizzas
Answer:
1048576
Step-by-step explanation:
Given the following :
Pizza order :
Size = small, medium, large, or extra large = 4 possible sizes
Toppings = any combination of 8 possible toppings (getting no toppings is allowed, as is getting all 8).
Combination of Toppings = 2^8
Four different sizes of pizza = 4
Number of possibilities in ordering for a single pizza :
(4 * 2^8) = 4 * 256 = 1024
Number of possibilities in ordering two pizzas :
(4 * 2^8)^2
(2^2 * 2^8)^2
From indices :
[2^(2+8)]^2
[2^(10)]^2
2^(10*2)
2^20
= 1048576
Robert is putting new roofing shingles on his house. Each shingle is 1 2/3 feet long. The north part of the house has a roof line that is 60 feet across. How many shingles can be placed (side by side) on the north part of the house?
Answer: 36 shingles can be placed on the north part of the house.
Step-by-step explanation:
Given: Length of each shingle = [tex]1\dfrac23[/tex] feet = [tex]\dfrac53[/tex] feet.
The north part of the house has a roof line that is 60 feet across.
Then, the number of shingles can be placed on the north part of the house = (Length of roof line in north part) ÷ (Length of each shingle)
[tex]=60\div \dfrac{5}{3}\\\\=60\times\dfrac{3}{5}\\\\=12\times3=36[/tex]
Hence, 36 shingles can be placed on the north part of the house.
Gail paid a total of $12,000 for stock that was $6 per share. If she sold all her shares for $18,000, how much profit on each share did she make?
A
$9
B
$3
С.
S2000
D
$6.000
Answer:
$3
Step-by-step explanation:
Given
Total Cost Price: $12,000
Unit Cost Price= $6
Total Selling Price = $18,000
Required
Determine the profit on each share
First, we need to determine the units of share bought;
Units = Total cost price / Unit Cost Price
[tex]Units = \frac{\$12000}{\$6}[/tex]
[tex]Units = 2000[/tex]
Next is to determine the selling price of each share; This is calculated as follows;
Unit Selling Price = Total Selling Price / Units Sold
[tex]Unit\ Selling\ Price = \frac{\$18000}{\$2000}[/tex]
[tex]Unit\ Selling\ Price = \$9[/tex]
The profit is the difference between the unit cost price and unit selling price
[tex]Profit = Unit\ Selling\ Price - Unit\ Cost\ Price[/tex]
[tex]Profit = \$9 - \$6[/tex]
[tex]Profit = \$3[/tex]
The cost, C, in United States Dollars ($), of cleaning up x percent of an oil spill along the Gulf Coast of the United States increases tremendously as x approaches 100. One equation for determining the cost (in millions $) is:
Complete Question
On the uploaded image is a similar question that will explain the given question
Answer:
The value of k is [tex]k = 214285.7[/tex]
The percentage of the oil that will be cleaned is [tex]x = 80.77\%[/tex]
Step-by-step explanation:
From the question we are told that
The cost of cleaning up the spillage is [tex]C = \frac{ k x }{100 - x }[/tex] [tex]x \le x \le 100[/tex]
The cost of cleaning x = 70% of the oil is [tex]C = \$500,000[/tex]
Now at [tex]C = \$500,000[/tex] we have
[tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]k = 214285.7[/tex]
Now When [tex]C = \$900,000[/tex]
[tex]x = 80.77\%[/tex]
During two years in college, a student earned $9,500. The second year she earned $500 more than twice the amount she earned the first year. How much did she earn the first year?
Find usubscript10 in the sequence -23, -18, -13, -8, -3, ...
Step-by-step explanation:
utilise the formula a+(n-1)d
a is the first number while d is common difference
Answer:
22
Step-by-step explanation:
Using the formular, Un = a + (n - 1)d
Where n = 10; a = -23; d = 5
U10 = -23 + (9)* 5
U10 = -23 + 45 = 22
