Answer:
5.94% of customers carries a balance of GH¢100 or lower.
82.64% of customers carries a balance of GH¢500 or lower.
0% of current account customers carries average daily balances exactly equal to GH¢500.
76.7% of customers maintains account balance between GH¢100 and GH¢500
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 350, \sigma = 160[/tex]
What percentage of customers carries a balance of GH¢100 or lower?
This is the pvalue of Z when X = 100. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{100 - 350}{160}[/tex]
[tex]Z = -1.56[/tex]
[tex]Z = -1.56[/tex] has a pvalue of 0.0594
5.94% of customers carries a balance of GH¢100 or lower.
What percentage of customers carries a balance of GH¢500 or lower?
This is the pvalue of Z when X = 500.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{500 - 350}{160}[/tex]
[tex]Z = 0.94[/tex]
[tex]Z = 0.94[/tex] has a pvalue of 0.8264
82.64% of customers carries a balance of GH¢500 or lower.
What percentage of current account customers carries average daily balances exactly equal to GH¢500?
In the normal distribution, the probability of finding a value exactly equal to X is 0. So
0% of current account customers carries average daily balances exactly equal to GH¢500.
What percentage of customers maintains account balance between GH¢100 and GH¢500?
This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 100.
From b), when X = 500, Z = 0.94 has a pvalue of 0.8264
From a), when X = 100, Z = -1.56 has a pvalue of 0.0594
0.8264 - 0.0594 = 0.767
76.7% of customers maintains account balance between GH¢100 and GH¢500
Find the domain and the range of the relation. Determine whether the relation is a function.
{(-1,0),(7,0),(-1,9),(-6,-9)}
Answer:
In a relation h(x) = y, the values of x are the domain and the values of y are the range.
Here the domain is:
D = {-6, -1, 7}
The range is:
R = {-9, 0 , 9}
Now, a relation f(x) = y is a function only if for every x in the domain we have only and only one value in the range.
Here we can see that for the value -1 in the domain we have two different values in the range:
(-1, 0) and (-1, 9)
So this can not be a function.
(If you want to take the variable as y, you also have that the value y = 0 leads to two different values in x, so this cant be a function either)
I NEED HELP WITH THIS PLEASE HELP ME
Answer:
156 minutes
Step-by-step explanation:
So we need to create an equation to represent how Frank's phone company bills him
I will denote "y" as the total for his billI will denote "x" as the number of minutes Frank usesSo the phone company charges an $8 monthly fee, so this value remains constant and will be our "y-intercept"
They then charge $0.06 for every minute he talks, this will be our "slope"
Combining everything into an equation, we have: y = 0.06x + 8
Now since we were given Franks phone bill total and want to figure out how many minutes he used, we just need to solve the equation for x and plug in our known y value
y = 0.06x + 8 → y - 8 = 0.06x → [tex]x=\frac{y-8}{0.06}[/tex] Then plugging in our y value we get [tex]x=\frac{17.36-8}{0.06}=\frac{9.36}{0.06}= 156[/tex]Frank used up a total of 156 minutes
What’s the correct answer for this question?
Answer: choice D 1/2
Step-by-step explanation:
Events A and B are independent if the equation P(A∩B) = P(A) · P(B) holds true.
so
1/6=1/3*p(A)
p(A)=1/2
Assume Shelley Kate decides to take her social security at age 63. What amount of social security benefit will she receive each month, assuming she is entitled to $720 per month
She will receive a lot more money because she is already retired from work already and will win as bit more money
Find the volume of a right circular cone that has a height of 8.1 in and a base with a diameter of 7.6 in. Round your answer to the nearest tenth of cubic inch
Answer:
[tex]V=122.5in^3[/tex]
Step-by-step explanation:
The volume of a right circular cone is given by:
[tex]V=\frac{\pi r^2h}{3}[/tex]
where r is the radius of the circle and h is the height of the cone, and [tex]\pi[/tex] is a constant [tex]\pi=3.1416[/tex].
According to the problem the height is:
[tex]h=8.1 in[/tex]
and we don't have the radius but we have the diameter, which is useful to find it. We just divide the diameter by 2 to find the radius:
[tex]r=\frac{d}{2}=\frac{7.6in}{3}=3.8in[/tex]
Now, we can find the volume by substituting all the known values:
[tex]V=\frac{\pi r^2h}{3}[/tex]
[tex]V=\frac{(3.1416)(3.8in)^2(8.1in)}{3} \\\\V=\frac{(3.1416)(14.44in^2)(8.1in)}{3} \\\\V=\frac{367.454in^3}{3} \\\\V=122.485[/tex]
Rounding the volume to the nearest tenth of cubic inch we get:
[tex]V=122.5in^3[/tex]
In a village
The number of houses and the number of flats are in the ratio 9:5
The number of flats and the number of bungalows are in the ratio 10:3
There are 30 bungalows in the village.
How many houses are there in the village?
Note: please make sure your final answer clear by writing ... houses
The number of houses are 180, and the number of flats are 100.
It is given that the number of houses and the number of flats ratio is 9:5 the number of flats and the number of bungalows ratio is 10:3.
It is required to find the number of houses in the village if the number of bungalows is 30.
What is a fraction?Fraction number consists of two parts one is the top of the fraction number which is called the numerator and the second is the bottom of the fraction number which is called the denominator.
The ratio of the number of houses and the number of flats:
= 9:5 and
The ratio of the number of flats and the number of bungalows :
=10:3
It means we can write the ratio of the number of houses and the number of flats = 18:10
And the ratio of the number of:
Houses : Flats : Bungalows = 18:10:3
But the number of bungalows are 30.
Then the ratios are:
180:100:30
Thus, the number of houses are 180, and the number of flats are 100.
Learn more about the fraction here:
brainly.com/question/1301963
Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications, which if a multiplication takes sec would take these times:
n 10 15 20 25
Time 0.004 sec 22 min 77 years 0.5.109years
Answer:
number of multiplies is n!n=10, 3.6 msn=15, 21.8 minn=20, 77.09 yrn=25, 4.9×10^8 yrStep-by-step explanation:
Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...
mpy[n] = n·mp[n-1]
mpy[2] = 2
So, ...
mpy[n] = n! . . . n ≥ 2
__
If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...
10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10
Then the larger matrices take ...
n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min
n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years
n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years
_____
For the shorter time periods (less than 100 years), we use 365.25 days per year.
For the longer time periods (more than 400 years), we use 365.2425 days per year.
For a super soaker water gun, a pump handle is moved back and forth to build up pressure in the water reservoir. The water is released by pulling a trigger and shooting the water a significant distance. Assuming that you can create an absolute pressure of 8 atm in the reservoir:
a) What is the velocity at which the water leaves the gun?
b) If the water exits the gun through a hole with a radius of 1-mm, what is the volume rate of flow in m3/s?
c) If the water gun is fired horizontally and held 1.2 meters above the ground, where does the water hit the ground? Pressure 8 cm water
Answer:
a) The velocity at which the water leaves the gun = 37.66 m/s
b) The volume rate of flow = (1.183 × 10⁻⁴) m³/s
c) The water hits the ground 18.64 m from the point where the water gun was shot.
Step-by-step explanation:
a) Using Bernoulli's equation, an equation that is based on the conservation of energy.
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
The two levels we are considering is just inside the water reservoir and just outside it.
ρgh is an extension of potential energy and since the two levels are at the same height,
ρgh₁ = ρgh₂
Bernoulli's equation becomes
P₁ + ½ρv₁² = P₂ + ½ρv₂²
P₁ = Pressure inside the water reservoir = 8 atm = 8 × 101325 = 810,600 Pa
ρ = density of water = 1000 kg/m³
v₁ = velocity iof f water in the reservoir = 0 m/s
P₂ = Pressure outside the water reservoir = atmospheric pressure = 1 atm = 1 × 101325 = 101,325 Pa
v₂ = velocity outside the reservoir = ?
810,600 + 0 = 101,325 + 0.5×1000×v₂²
500v₂² = 810,600 - 101,325 = 709,275
v₂² = (709,275/500) = 1,418.55
v₂ = √(1418.55) = 37.66 m/s
b) Volumetric flowrate is given as
Q = Av
A = Cross sectional Area of the channel of flow = πr² = π×(0.001)² = 0.0000031416 m²
v = velocity = 37.66 m/s
Q = 0.0000031416 × 37.66 = 0.0001183123 m³/s = (1.183 × 10⁻⁴) m³/s
c) If the height of gun above the ground is 1.2 m. Where does the water hit the ground?
The range of trajectory motion is given as
R = vT
v = horizontal component of the velocity = 37.66 m/s
T = time of flight = ?
But time of flight is given as
T = √(2H/g) (Since the initial vertical component of the velocity = 0 m/s
H = 1.2 m
g = acceleration due to gravity = 9.8 m/s²
T = √(2×1.2/9.8) = 0.495 s
Range = vT = 37.66 × 0.495 = 18.64 m
Hope this Helps!!!
A man driving a car leaves a point A drives up to 32.5 km in a direction of 070. A cyclist leaves the same point in a direction 130 travelling. After some few hours both drivers are 80 km apart. Use this information to answer 3 questions. (1). What is the distance covered by the cyclist at this time in 2 d.p. (2). Find the bearing of Cyclist from the Car. correct to 1 d.p. (3). Find the shortest distance between the car and the line of path of the cyclist, in 2 d.p.
Answer: No 1 is 91.14 km who else could help with the rest of the solution for number 1, 2 & 3.
if you’re good with permutations in math 30 help out with this easy question
In how many ways can five boys and three girls sit in a row such that all boys sit together?
a) 4800
b) 5760
c) 2880
d) 1440
Answer:
2880
Step-by-step explanation:
Consider the 5 boys to be 1 group. The boys and 3 girls can be arranged in 4! ways.
Within the group, the boys can be arranged 5! ways.
The total number of permutations is therefore:
4! × 5! = 2880
Simplify this equation x2-5x-36
Answer:
[tex]=\left(x+4\right)\left(x-9\right)[/tex]
Step-by-step explanation:
[tex]x^2-5x-36\\\mathrm{Break\:the\:expression\:into\:groups}\\=\left(x^2+4x\right)+\left(-9x-36\right)\\\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2+4x\mathrm{:\quad }x\left(x+4\right)\\\mathrm{Factor\:out\:}-9\mathrm{\:from\:}-9x-36\mathrm{:\quad }-9\left(x+4\right)\\=x\left(x+4\right)-9\left(x+4\right)\\\mathrm{Factor\:out\:common\:term\:}x+4\\=\left(x+4\right)\left(x-9\right)[/tex]
please hurry I’ll make brainiest
A marble is thrown off of a balcony, towards the ground, from a height
of 18 feet above ground level, with a velocity of 4.5 feet per second.
Which function could be used to model the height of the marble, after
t seconds?
Answer:
Option (3)
Step-by-step explanation:
A stone has been thrown off towards the ground from a height [tex]h_{0}[/tex] = 18 feet
Initial speed of the stone 'u' = 4.5 feet per second
Since height 'h' of a projectile at any moment 't' will be represented by the function,
h(t) = ut - [tex]\frac{1}{2}(g)(t)^2[/tex] + [tex]h_{0}[/tex]
h(t) = 4.5t - [tex]\frac{1}{2}(32)t^2[/tex]+ 18 [ g = 32 feet per second square]
h(t) = 4.5t - 16t² + 18
h(t) =-16t² + 4.5t + 18
Therefore, Option (3) will be the answer.
Results of 99% confidence intervals are consistent with results of two-sided tests with which significance level? Explain the connection. A 99% confidence interval is consistent with a two-sided test with significance level alphaequals nothing because if a two-sided test with this significance level does not reject the null hypothesis, then the confidence interval ▼ contains does not contain the value in the null hypothesis.
Answer:
Yes, they are consistent.
A 99% confidence interval is consistent with a two-sided test with significance level alpha=0.01 because if a two-sided test with this significance level does not reject the null hypothesis, then the confidence interval does contains the value in the null hypothesis.
Step-by-step explanation:
Yes, they are consistent.
A 99% confidence interval is consistent with a two-sided test with significance level alpha=0.01 because if a two-sided test with this significance level does not reject the null hypothesis, then the confidence interval does contains the value in the null hypothesis.
The critical values of the confidence level are equivalent to the critical values in the hypothesis test. In the case that the conclusion of the test is to not reject the null hypothesis, the test statistic falls within the acceptance region: its value is within the critical values of the two-sided test.
Then, it is also within the critical values of the confidence interval and the sample mean (or other measure) will be within the confidence interval bounds.
At a gas station, 50% of the customers use regular gas, 30% use mid-grade gas and 20% use premium gas. Of those customers using regular gas, only 30% fill their tanks. Of those customers using mid-grade gas, 60% fill their tanks, whereas of those using premium, 50% fill their tanks. What is the probability that the next customer will request mid-grade gas and fill the tank
Answer:
The probability that the next customer will request mid-grade gas and fill the tank is 0.1800
Step-by-step explanation:
In order to calculate the probability that the next customer will request mid-grade gas and fill the tank we would have to make the following calculation:
probability that the next customer will request mid-grade gas and fill the tank= percentage of the people using mid-grade gas* percentage of the people using mid-grade gas that fill their tanks
probability that the next customer will request mid-grade gas and fill the tank= 30%*60%
probability that the next customer will request mid-grade gas and fill the tank= 0.1800
The probability that the next customer will request mid-grade gas and fill the tank is 0.1800
In a grinding operation, there is an upper specification of 3.150 in. on a dimension of a certain part after grinding. Suppose that the standard deviation of this normally distributed dimension for parts of this type ground to any particular mean dimension LaTeX: \mu\:is\:\sigma=.002 μ i s σ = .002 in. Suppose further that you desire to have no more than 3% of the parts fail to meet specifications. What is the maximum (minimum machining cost) LaTeX: \mu μ that can be used if this 3% requirement is to be met?
Answer:
Step-by-step explanation:
Let X denote the dimension of the part after grinding
X has normal distribution with standard deviation [tex]\sigma=0.002 in[/tex]
Let the mean of X be denoted by [tex]\mu[/tex]
there is an upper specification of 3.150 in. on a dimension of a certain part after grinding.
We desire to have no more than 3% of the parts fail to meet specifications.
We have to find the maximum [tex]\mu[/tex] such that can be used if this 3% requirement is to be meet
[tex]\Rightarrow P(\frac{X- \mu}{\sigma} <\frac{3.15- \mu}{\sigma} )\leq 0.03\\\\ \Rightarrow P(Z <\frac{3.15- \mu}{\sigma} )\leq 0.03\\\\ \Rightarrow P(Z <\frac{3.15- \mu}{0.002} )\leq 0.03[/tex]
We know from the Standard normal tables that
[tex]P(Z\leq -1.87)=0.0307\\\\P(Z\leq -1.88)=0.0300\\\\P(Z\leq -1.89)=0.0293[/tex]
So, the value of Z consistent with the required condition is approximately -1.88
Thus we have
[tex]\frac{3.15- \mu}{0.002} =-1.88\\\\\Rrightarrow \mu =1.88\times0.002+3.15\\\\=3.15[/tex]
SELECT THE EQUIVALENT EXPRESSION
(6^-4 x 8^-7)^-9
A. 6^36•8^63
B. 1/6^13•8^16
Answer:
A
Step-by-step explanation:
Calculate the products in the multiple choice and see if any equal the product in the problem.
Hence as the products calculated in choice A equal that in the problem;the answer is A
Please help! Will give Brainliest!
Steps 1-4 in attachment (#4 below)
Step 4: Use the equation you wrote in Step 3. Write the equation for the graph of g(x) that has also been shifted right 1 unit.
Answer:
g(x) = 2|x|g(x) = -2|x|g(x) = -2|x| -3g(x) = -2|x-1| -3Step-by-step explanation:
1) Vertical stretch is accomplished by multiplying the function value by the stretch factor. When |x| is stretched by a factor of 2, the stretched function is ...
g(x) = 2|x|
__
2) Reflection over the x-axis means each y-value is replaced by its opposite. This is accomplished by multiplying the function value by -1.
g(x) = -2|x|
__
3) As you know from when you plot a point on a graph, shifting it down 3 units subtracts 3 from the y-value.
g(x) = -2|x| -3
__
4) A right-shift by k units means the argument of the function is replaced by x-k. We want a right shift of 1 unit, so ...
g(x) = -2|x -1| -3
Find the focus and directrix of the parabola y = {(x + 2)2 – 3.
Answer:
focus: (-2, -2.75)directrix: y = -3.25Step-by-step explanation:
For focus-to-vertex distance "p", the equation of a parabola with vertex (h, k) can be written as ...
y = 1/(4p)(x -h)^2 +k
Comparing this to your equation, we see that ...
1/(4p) = 1
h = -2
k = -3
Solving for p, we find ...
1/(4p) = 1
1/4 = p . . . . . multiply by p
The parabola opens upward, so this means the focus is 1/4 unit above the vertex, and the directrix is 1/4 unit below the vertex.
focus: (-2, -2.75)directrix: y = -3.25Answer:
The focus is at (–2,–212) and the directrix is at y = –312.
Step-by-step explanation:
Find the focus and directrix of the parabola y=12(x+2)2−3.
got the answer right in the test.
John Calipari, head basketball coach for the national champion University of Kentucky Wildcats, is the highest paid coach in college basketball with an annual salary of million (USA Today, March 29, 2012). The following sample shows the head basketball coach's salary for a sample of schools playing NCAA Division basketball. Salary data are in millions of dollars.
University Coach's Salary University Coach's Salary
Indian 2.2 Syracuse 1.5
Xavier .5 Murry State .2
Texas 2.4 Florida State 1.1
Connecticut 2.7 South Dekota State .1
West Virginia 2.0 Vermont .2
A. Use the sample mean for the 10 schools to estimate the population mean annual salary for head basketball coaches at colleges and universities playing NCAA Division 1 basketball (to 2 decimal).
B. Use the data to estimate the population standard deviation for the annual salary for head basketball coaches (to 4 decimals).
C. what is the 95% confidence interval for the population variance (to 2 decimals)?
D. what is the 95% confidence interval for the population standard deviation (to 2 decimals)?
Answer:
A) sample mean = $1.36 million
B) standard deviation = $0.9189 million
C) variance confidence interval = ($0.40 million, $2.81 million)
D) standard deviation confidence interval = ($1.93 million , $0.79 million)
*since the sample size is very small, the confidence interval is not valid.
Step-by-step explanation:
samples:
$2.7 million$2.4 million$2.2 million$2 million$1.5 million$1.5 million$0.5 million$0.5 million$0.2 million$0.1 millionsample mean = $1.36 million
the standard deviation:
$2.7 million - $1.36 million = 1.34² = 1.7956$2.4 million - $1.36 million = 1.04² = 1.0816$2.2 million - $1.36 million = 0.84² = 0.7056$2 million - $1.36 million = 0.64² = 0.4096$1.5 million - $1.36 million = 0.14² = 0.0196$1.5 million - $1.36 million = 0.14² = 0.0196$0.5 million - $1.36 million = -0.86² = 0.7396$0.5 million - $1.36 million = -0.86² = 0.7396$0.2 million - $1.36 million = -1.16² = 1.3456$0.1 million - $1.36 million = -1.26² = 1.5876total $8.444 million / 10 = $0.8444 millionvariance 0.8444
standard deviation = √0.8444 = 0.9189
in order to calculate the confidence interval for the population variance we are going to use a chi-square distribution with 2.5% on each tail ⇒ table values 2.7004 and 19.023 enclose 95% of the distribution.
[(n - 1) x variance] / 2.7004 = (9 x 0.8444) / 2.7004 = 2.81
[(n - 1) x variance] / 19.023 = (9 x 0.8444) / 19.023 = 0.40
95% confidence interval = mean +/- 1.96 standard deviations/√n:
$1.36 million + [(1.96 x $0.9189 million)/√10] = $1.36 million + $0.57 million = $1.93 million
$1.36 million - $0.57 million = $0.79 million
if y=5x what happens to the value of y if the value of x doubles
Answer:
[tex] y = 5x[/tex]
And we need to ee what happen if we increase the value of x by a factor of 2. So then for this case we can set up the equation like this:
[tex] y_f = 5(2x) = 10x[/tex]
And if we find the ratio between the two equations we got:
[tex] \frac{y_f}{y} =\frac{10x}{5x} =2[/tex]
So then if we increase the value of x by a factor of 2 then the value of y increase also by a factor of 2
Step-by-step explanation:
For this case we have this equation given:
[tex] y = 5x[/tex]
And we need to ee what happen if we increase the value of x by a factor of 2. So then for this case we can set up the equation like this:
[tex] y_f = 5(2x) = 10x[/tex]
And if we find the ratio between the two equations we got:
[tex] \frac{y_f}{y} =\frac{10x}{5x} =2[/tex]
So then if we increase the value of x by a factor of 2 then the value of y increase also by a factor of 2
Will pick brainliest! I need help with this, actual effort in answering is much appreciated.
Answer:
option 2
Step-by-step explanation:
4^2=16/8=2. 4^2=16/16=1. 2-1=1
Ellen is opening. Cookie shop eat a local middle school . She randomly surgery’s students to determine The types of cookies they would buy at the cookie shop the results of the surveys are below based on the survey which statement is not true
Complete Question:
Ellen is opening. Cookie shop eat a local middle school . She randomly surveys students to determine The types of cookies they would buy at the cookie shop the results of the surveys are below based on the survey. Which statement is not true?
A) If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.
B) 22½% of the students surveyed would buy a sugar cookie.
C) If 240 students were to purchase a cookie from the store, approximately 66 students will purchase a peanut butter cookie.
D) Half of the students prefer chocolate chips or oatmeal raisin cookies.
(Table showing the results of her survey is in the attachment below)
Answer:
A) If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.
Step-by-step Explanation:
STEP 1:
In order to ascertain which of the statements given in the options that is NOT TRUE, let's express the given number of students that would buy the different types of cookies in their PROPORTIONS AND PERCENTAGES in the survey result.
Thus:
Type of Cookie==> no of students => Proportion (no of students of a cookie type ÷ total no of students) => % (each proportion for a cookie type)
Chocolate Chip==> 25 => 0.3125 (25÷80) => 31.25% (0.3125 × 100)
Oatmeal Raisin==> 15 => 0.1875 (15÷850) => 18.75% (0.1875 × 100)
Peanut Butter== 22 => 0.275 (22÷80) => 27.5% (0.275 × 100)
Sugar ==> 18 => 0.225 (18÷80) => 22.5% (0.225 × 100)
STEP 2:
Next step is to consider each statement given in the question to see if they are true or not.
==>OPTION A: If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.
To find out if this is true, multiply the proportion of students who would be buy a sugar cookie (0.255) by 160 = 0.255 × 160 = 36.
If 160 buy a cookie, approximately 36 students would buy a sugar.
Option A IS NOT TRUE.
OPTION B: 22½% of the students surveyed would buy a sugar cookie.
From our calculation in STEP 2, we have:
Sugar ==> 18 => 0.225 (18÷80) => 22.5% (0.225 × 100).
Option B is TRUE. 22.5% of the students would go for sugar cookie.
OPTION C: If 240 students were to purchase a cookie from the store, approximately 66 students will purchase a peanut butter cookie.
Number of students that would buy the peanut butter if 240 students were to get a cookie = proportion of students that opt for peanut butter cookie in the survey (0.275) × 240 = 0.275 × 240 = 66.
Option C is TRUE.
OPTION D: D) Half of the students prefer chocolate chips or oatmeal raisin cookies.
Proportion of students who prefer chocolate chips or oatmeal raisin cookies = 0.3125 + 0.1785 = 0.491
This is approximately 0.5 = ½ of the total number of students.
Option D is TRUE.
Therefore, we can conclude that the statement, "A) If 160 students buy a cookie, approximately 18 students would buy a sugar cookie" is NOT TRUE.
PLS HELP ME 10PTS
An artist creates a cone-shaped sculpture for an art exhibit. If the sculpture is 7 feet tall and has a base with a circumference of 27.632 feet, what is the volume of the sculpture?
Answer: The volume of the sculpture is 141.84 cubic-feet
Step-by-step explanation: Please see the attachments below
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Solution,
Radius=2 m
Area =pi r^2
= 3.142*(2)^2
=12.568 m^2
hope it helps
Good luck on your assignment
On a number line, b, is located the same distance from 0 as another number, a, but in the opposite direction. The number b varies directly with number a. For example b= 11/4 when a= -11/4
A) b=-a
B) -b=-a
C) b-a=0
D) b(-a)=0
Answer:
B and A
Step-by-step explanation:
So based on the facts given, we know that b and a both have the same abasolute value. It does not matter whether a or b is negative or positive.
Each limit represents the derivative of some function f at some number a. State such an f and a in each case.
lim √9 + h - 3 / h
h-->0
Answer:
a = 0f(h) = [tex]\frac{\sqrt{9+h} - 3}{h}[/tex]limit of the function is 1/6Step-by-step explanation:
The general form representing limit of a function is expressed as shown below;
[tex]\lim_{h \to a} f(h)[/tex] where a is the value that h will take and use in the function f(h). It can be expressed in words as limit of function f as h tends to a. Comparing the genaral form of the limit to the limit given in question [tex]\lim_{h \to 0} \frac{\sqrt{9+h} - 3}{h}[/tex], it can be seen that a = 0 and f(h) = [tex]\frac{\sqrt{9+h} - 3}{h}[/tex]
Taking the limit of the function
[tex]\lim_{h \to 0} \frac{\sqrt{9+h} -3}{h}\\= \frac{\sqrt{9+0}-3 }{0}\\= \frac{0}{0}(indeterminate)[/tex]
Applying l'hopital rule
[tex]\lim_{h \to 0} \frac{\frac{d}{dh} (\sqrt{9+h} - 3)} {\frac{d}{dh} (h)}\\= \lim_{h \to 0} \frac{1}{2} (9+h)^{-1/2} /1\\=\frac{1}{2} (9+0)^{-1/2}\\= \frac{1}{2} * \frac{1}{\sqrt{9} } \\= 1/2 * 1/3\\= 1/6[/tex]
Please answer this question for me thank you !! 20 Points !! Will give brainliest !!
Answer:
b
Step-by-step explanation:
In a parralel graph, the slopes would always be the same. The intercept in the answer is 2, showing that the coordinate points are (0,2)
Hope this helps!:)
Answer:
B) y = 2x + 2
Step-by-step explanation:
Firstly, you have to know that parallel lines have congruent slopes. That means that the slope of this line will be 2.
Next, make a point slope form of the equation:
y - y1 = m(x - x1)
y - 2 = 2(x - 0)
y - 2 = 2x - 0
Now, we can make it into slope intercept form.
y - 2 = 2x
y = 2x + 2
Hope this helps :)
Function c(x) = 5x
If your input was 2, what is your output?
How do I set up this problem. I'm lost
Answer:
the answer is 64 .
Step-by-step explanation:
basically i just divided 48 by 2.4 and got 20 .. so that means that 20 has to be the multiplied factor so i just multiplied 3.2 by 20 and got 64.
(Bonus) A rectangular box has its edges changing length as time passes. At a par-ticular instant, the sides have lengthsa= 150 feet,b= 80 feet, andc= 50 feet.At that instant,ais increasing at 100 feet/sec,bis decreasing 20 feet/sec, andcisincreasing at 5 feet/sec. Determine if the volume of the box is increasing, decreasing,or not changing at all, at that instant.
Answer:
the volume of the box is increasing
dV = +310,000 ft^3/s
Step-by-step explanation:
Volume of a rectangular box with side a,b and c can be expressed as;
V = abc
The change in volume dV can be expressed as;
dV = d(abc)/da + d(abc)/db + d(abc)/dc
dV = bc.da + ac.db + ab.dc ......1
Given:
a= 150 feet,
b= 80 feet, and
c= 50 feet
ais increasing at 100 feet/sec,bis decreasing 20 feet/sec, andcisincreasing at 5 feet/sec
da = +100 feet/s
db = -20 feet/s
dc = +5 feet/s
Substituting the values into the equation 1;
dV = (80×50×+100) + (150×50×-20) + (150×80×+5)
dV = +400000 - 150000 + 60000 ft^3/s
dV = +310,000 ft^3/s
Since dV is positive, the volume of the box is increasing at that instant.