Answer:
0.0908 [tex]m^{2}[/tex] (to 3 S.F.)
Step-by-step explanation:
Area = π[tex]r^{2}[/tex]
π * [tex]17^{2}[/tex] = 907.92
= 908 [tex]cm^{2}[/tex]
=0.0908 [tex]m^{2}[/tex]
What is the value of -(3/4) to the power of -4
The answer would be -3 13/81 (simplified)
What would be the approximate 95% confidence interval for the mean number of ounces of catchup bottle in the sample
Answer:
The 95% confidence interval for the mean number of ounces of ketchup bottle is (23.8, 24.2).
Step-by-step explanation:
The complete question is:
Suppose that a restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average, with a standard deviation of 0.8 ounces. If you took a sample of 49 bottles of ketchup, what would be the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample?
Solution:
The (1 - α)% confidence interval for the population mean is:
[tex]CI=\bar x\pm z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}[/tex]
The information provided is:
[tex]\bar x=24\\\sigma=0.8\\n=49\\\text{Confidence Level}=95\%[/tex]
The critical value of z for 95% confidence level is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
*Use a z-table.
Compute the 95% confidence interval for the mean number of ounces of ketchup per bottle as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}[/tex]
[tex]=24\pm1.96\cdot \frac{0.80}{\sqrt{49}}\\\\=24\pm 0.224\\\\=(23.776, 24.224)\\\\\approx (23.8, 24.2)[/tex]
Thus, the 95% confidence interval for the mean number of ounces of ketchup bottle is (23.8, 24.2).
Labrador Retriever weighs 48 kg after a diet and exercise program the dog weighs 43 kilograms to determine if this shows a percent increase or decrease and explain why what is the percent change of its weight a 10% B 11% C 110% D 111% please help.
Answer:
percentage change in weight ≈ 10%
Step-by-step explanation:
The dog weighed 48 kg after a diet and after an exercise program the dog had a weight of 43 kg. This means the dog loss weight since the dog weight decreased from an initial value of 48 kg to 43 kg. The decrease in weight can be calculate as
decrease in weight = original weight - new weight
original weight = 48 kg
new weight = 43 kg
decrease in weight = 48 - 43 = 5 kg
Since the weight decrease their will be a percentage decrease in weight.
% decrease = decrease in weight/original weight × 100
% decrease = 5/48 × 100
% decrease = 500/48
% decrease = 10. 42666666667
percentage change in weight ≈ 10%
Classify the following triangle .check all that apply
Answer:
Its right and scalene.
It has a right angle and all the sides are diferent.
Find the equation of a line perpendicular to 2x-4y=1 that contains the point (-4, -2).
Answer:
y = -2x - 10
Step-by-step explanation:
1. rearrange to find the gradient.
the gradient of the original equation is 1/2 hence why a line PERPENDICULAR to that equation would have a gradient of -2.
2. substitute into y - y1 = m (x - x1)
y - (-2) = -2 (x - (-4))
y = -2x - 10
An icicle it's in the shape of an inverted cone with a diameter of 12 m m and a height of 100 mm. How much Frozen water is in the icicle? Round to the nearest hundredth.
Answer:
3768 mm ^ 3
Step-by-step explanation:
We have that the volume of a cone is given by:
V = 1/3 * Ac * h
Where Ac is the area of the circle, we know that the radius is half the diameter then:
r = d / 2
r = 12/2
r = 6
And Ac is equal to:
Ac = pi * r ^ 2
replacing:
Ac = 3.14 * 6 ^ 2
Ac = 113.04
113.04 mm ^ 2 is the area of the circle, replacing the volume form knowing that h = 100
V = 1/3 * 113.04 * 100
V = 3768
Therefore they fit a total of 3768 mm ^ 3
Find the area of a circle with radius, r = 6.89m.
Give your answer rounded to 2 DP (2 decimal points)
The photo is attached below
Answer:
149.14 [tex]m^{2}[/tex]
Step-by-step explanation:
Area of a circle = π[tex]r^{2}[/tex]
so A = π * 6.89^2 = 149.14 (to 2d.p.)
Which sequences are geometric? Check all that apply.
O 1,5, 25, 125, ...
3, 6, 9, 12,...
3, 6, 12, 24, ...
3, 9, 81, 6, 561, ...
10, 20, 40, 60, ...
Answer:
1, 5, 25, 125, ...
3, 6, 12, 24, ...
Step-by-step explanation:
a geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio1, 5, 25, 125, ...
yes, the common ratio is 53, 6, 9, 12,...
no3, 6, 12, 24, ...
yes, the common ratio is 23, 9, 81, 6, 561, ...
no10, 20, 40, 60, ...
noIf the area of a triangle is 36 in.^2in. 2 and the base is 9 in., what is the height of the triangle?
Answer:
Height = 8
Step-by-step explanation:
Area of a triangle = [tex]\frac{Base*Height}{2}[/tex]
Say the height = x
4.5x = 36
x = 8
fand f are functions.
If f(4) = 2 then f'(2) = ?
Answer:
4
Step-by-step explanation:
[tex] \because \: f(4) = 2 \\ \therefore \: {f}^{ - 1} (f(4)) = {f}^{ - 1} (2) \\ \therefore \: 4 = {f}^{ - 1} (2) \\ \huge \red{ \boxed{{f}^{ - 1} (2) = 4}}[/tex]
Answer:
4 four
Step-by-step explanation:
hope it helps you
At a football game, a vender sold a combined total of 152 sodas and hot dogs. The number of sodas sold was three times the number of hot dogs sold. Find the number of sodas and the number of hot dogs sold.
Answer:
38 hot dogs114 sodasStep-by-step explanation:
Sometimes problems of this nature are easily worked by considering groups of items. Here, it is convenient to consider a group as 1 hot dog and 3 sodas, so the number of sodas in the group is 3 times the number of hotdogs in the group.
Each group is 4 items, so 152/4 = 38 groups were sold.
In the 38 groups, there were 38 hot dogs and 3×38 = 114 sodas.
114 sodas and 38 hot dogs were sold.
Find the equation for the plane through the points Upper P 0 (5 comma 4 comma 5 ), Upper Q 0 (negative 5 comma negative 1 comma negative 4 ), and Upper R 0 (negative 2 comma 1 comma negative 2 ). The equation of the plane is nothing.
Answer:
The equation of the plane is
8(x - 5) - 7(y - 4) - 5(z - 5) = 0
8x - 7y - 5z + 13 = 0
Step-by-step explanation:
Given 3 points, P(x₁, y₁, z₁), Q(x₂, y₂, z₂), and R(x₃, y₃, z₃).
We can calculate the equation of the plane through those points as
a(x - x₀) + b(y - y₀) + c(z - z₀) = 0, where (x₀, y₀, z₀) are the coordinates of any one of the points P, Q, or R, and <a,b,c> is a vector perpendicular to the plane.
The vector perpendicular to the plane is obtained by writing vector PQ and PR and taking the cross or vector product.
For this question,
P = (5, 4, 5)
Q = (-5, -1, -4)
R = (-2, 1, -2)
PQ = (-5, -1, -4) - (5, 4, 5) = (-10, -5, -9)
= (-10î - 5ĵ - 9ķ)
PR = (-2, 1, -2) - (5, 4, 5) = (-7, -3, -7)
= (-7î - 3ĵ - 7ķ)
PQ × PR is then
| î ĵ ķ |
|-10 -5 -9|
|-7 -3 -7|
= î [(-5×-7) - (-9×-3)] - ĵ [(-10×-7) - (-9×-7)] + ķ [(-10×-3) - (-7×-5)]
= î (35 - 27) - ĵ (70 - 63) + ķ (30 - 35)
= 8î - 7ĵ - 5ķ
Hence, (a, b, c) = (8, -7, -5)
And using point P as (x₀, y₀, z₀) = (5, 4, 5)
The equation of the plane is
a(x - x₀) + b(y - y₀) + c(z - z₀) = 0
8(x - 5) - 7(y - 4) - 5(z - 5) = 0
8x - 40 - 7y + 28 - 5z + 25 = 0
8x - 7y - 5z = 40 - 28 - 25 = -13
8x - 7y - 5z + 13 = 0
Hope this Helps!!!
evaluate the formula of A=lw, for l=10.8 cm and w=2.5 cm
Answer:
A = 27 cm²
Step-by-step explanation:
[tex]A = lw\\Where, l=10.8 cm , w = 2.5 cm\\[/tex]
Putting in the above formula
A = (10.8)(2.5)
A = 27 cm²
A school needs 1,860 pencils for its students. The pencils are sold in boxes of 12. How many boxes does the school need to order?
Answer:
Step-by-step explanation:
155
The number of boxes required by the school to order is 155.
What is the unitary method?The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value. If an event can occur in m different ways and if following it, a second event can occur in n different ways, then the two events in succession can occur in m × n different ways.
We have been given that the school needs 1,860 pencils for its students. Also, the pencils are sold in boxes of 12.
We need to find the school needs to requires boxes to order.
Total number of pencil = 1,860
Number of boxes = 12
Therefore, boxes needed = 1,860 / 12
= 155
Hence, the number of boxes required by the school to order is 155.
To learn more about the unitary method, please visit the link given below;
https://brainly.com/question/23423168
#SPJ5
Christine has a motion detector light which gets activated an average of 16 times every 2 hours during the night. In order to find the probability that the motion detector light will be activated more than 4 times in a 25 minute period during the night using the Poisson distribution, what is the average number of activations per 25 minutes?
Answer:
The average number of activations per 25 minutes is 2.5.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!} [/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
In this question:
For 2 hours = 2*60 = 120 minutes, the mean is 16 times.
To find for 25 minutes, we use a rule of three.
120 minutes - 16 times
25 minutes - m times
[tex]120m = 12*25[/tex]
[tex]m = \frac{12*25}{120}[/tex]
[tex]m = 2.5[/tex]
The average number of activations per 25 minutes is 2.5.
Write an equation in slope-intercept form for the line that passes through (0,1) and (1,3)
Answer:
y= 2x+1
Step-by-step explanation:
Points:
(0,1) and (1,3)Form of the line:
y=mx+b, m- the slope, b- y-interceptFinding the slope:
m= (y2-y1)/(x2-x1)m=(3-1)/(1-0)= 2/1= 2Line is now:
y= 2x+bUsing one of the given points to find out the value of b:
1=2*0+bb=1So the equation for the line is:
y= 2x+1The illustration below is an example of a semi regular tessellation
true or false?
The correct answer is False
Explanation:
A tessellation refers to a regular pattern created by using regular polygons; additionally, in a tessellation, there are no gaps or spaces between the polygons. Besides this, a tesselation is categorized as regular if there is only one type of polygon in all the pattern or as semi-regular if there are two or more polygons but these still form a regular pattern. According to this, the illustration below is not a semi-regular tessellation because this only includes one polygon (hexagons), and therefore this would be classified as a regular tessellation.
Answer:
B
Step-by-step explanation:
In a survey of students, 60% were in high school and 40% were in middle school. Of the high school students, 30% had visited a foreign country. If a surveyed student is selected at random, what is the probability that the student is in high school and has visited a foreign country?
Answer:
The probability that the student is in high school and has visited a foreign country is 0.18.
Step-by-step explanation:
We are given that in a survey of students, 60% were in high school and 40% were in middle school.
Of the high school students, 30% had visited a foreign country.
Let the Probability that students were in high school = P(H) = 60%
Probability that students were in middle school = P(M) = 40%
Also, let F = event that students had had visited a foreign country
So, Probability that high school students had visited a foreign country = P(F/H) = 30%
Now, probability that the student is in high school and has visited a foreign country is given by = Probability that students were in high school [tex]\times[/tex] Probability that high school students had visited a foreign country
= P(H) [tex]\times[/tex] P(F/H)
= 0.60 [tex]\times[/tex] 0.30
= 0.18 or 18%
Solve 2x - 11 = k for x.
what is 0.84÷3 matggggg
[tex]0.84 \div 3 = 84 \div 300 = 0.28[/tex]
Which steps can be used to solve for the value of y?
(2013
(y +57) = 178
Answer: [tex]y = 121[/tex]
[tex](y+57) = 178[/tex]
[tex]y+57= 178[/tex]
[tex]y = 178 -57[/tex]
[tex]y = 121[/tex]
If a tank holds 4500 gallons of water, which drains from the bottom of the tank in 50 minutes, then Toricelli's Law gives the volume V of water remaining in the tank after t minutes as V = 4500 1 − 1 50 t 2 0≤ t ≤ 50. Find the rate at which water is draining from the tank after the following amounts of time.a) 5 min 855 x gal/min b) 10 min 160 x gal/min c) 20 min 120 x gal/min d) 50 min gal/min
Answer:
a) at 5 minutes: 162 gal/min
b) at 10 minutes: 144 gal/min
c) at 20 minutes: 108 gal/min
d) at 50 minutes: 0 gal/min
Step-by-step explanation:
Considering the formula given by the volume of water remaining in the tank:
[tex]V(t)=4500\,(1-\frac{1}{50} \,t)^2[/tex]we can find the rate of water draining from the tank, (that is change in volume divided elapsed time) with the derivative of the function at the different times. Notice that this function has a decaying curvature (see attached image) of volume as a function of time, and the idea is therefore to find the slope of the tangent line at the different requested times.
So we first calculate the derivative of this function at any time 't":
[tex]V(t)=4500\,(1-\frac{1}{50} \,t)^2\\V'(t)=9000\,(1-\frac{1}{50} \,t)\,(-\frac{1}{50})\\V'(t)=-180(1-\frac{1}{50} \,t)\\V'(t)=-180+3.6\,t[/tex]
And now we estimate this derivative at the different requested points for time values:
a) at 5 minutes: [tex]V'(5)=-180+3.6\,(5) = -162\,\,gal/min[/tex]
b) at 10 minutes: [tex]V'(10)=-180+3.6\,(10) = -144\,\,gal/min[/tex]
c) at 20 minutes: [tex]V'(20)=-180+3.6\,(20) = -108\,\,gal/min[/tex]
d) at 50 minutes: [tex]V'(50)=-180+3.6\,(50) = 0\,\,gal/min[/tex]
All the negative signs preceding indicate that the remaining volume in the tank is reducing as time goes by, so the volume at which the water is draining is actually the absolute value of those numbers.
find the circumference of the circle use 3.14 for pi when the radius is 13 cm
Answer:
C =81.64 cm
Step-by-step explanation:
The circumference of a circle is given by
C = 2*pi*r
C = 2 * 3.14 * 13
C =81.64 cm
_______________________________
Radius(r)=13 cm
Circumference of circle=?
Now,
Circumference of circle=2 pi r
=2*3.14*13
=81.64 cm
Hope it helps..
Good luck on your assignment
________________________________
The math department faculty at a large university wanted to know what portion of the student body believes students should be able to enroll in any math class without meeting a prerequisite. The statistics department offered to cooperate in conducting a survey, and a simple random sample of 500 students was selected from all the students enrolled in statistics classes. A survey form was sent by email to these 500 students and 236 responded. What is the population of interest for this study?
Answer:
The population of interest for this student is the students whom are enrolled in statistics classes.
Step-by-step explanation:
Sampling
This is a common statistics practice, when we want to study something from a population, we find a sample of this population.
For example:
I want to estimate the proportion of New York state residents who are Buffalo Bills fans. So i ask, lets say, 1000 randomly selected New York state residents wheter they are Buffalo Bills fans, and expand this to the entire population of New York State residents.
The population of interest are all residents of New York State.
A simple random sample of 500 students was selected from all the students enrolled in statistics classes.
This means that the population of interest for this student is the students whom are enrolled in statistics classes.
Find the first 4 terms and the 10th one n+5
Answer: First 4 terms of n + 5 = 6,7,8,9
10th term = 15
Hope this is right
Step-by-step explanation:
By putting n = 1 , 2, 3 , 4 we can find first 4 terms
When n = 1
n + 5 = 1 + 5 = 6
When n = 2
n + 5 = 2 + 5 = 7
When n = 3
n + 5 = 3 + 5 = 8
When n = 4
n + 5 = 4 + 5 = 9
When n = 10
n + 5 = 10 +5 = 15
The price of a visit to the dentist is $50. If the dentist fills any cavities, an additional charge of $100 per cavity
gets added to the bill.
Answer:
Cost of visit = 50 + 100n
Step-by-step explanation:
$50 is the set price of the visit.
$100 is the cost per cavity.
N is the number of cavities.
Since we don't know the number of cavities, 'n' will fill that spot.
100 x n will be the total cavity cost.
Cavity cost + set price of visit will equal the total cost of the visit.
Mist (airborne droplets or aerosols) is generated when metal-removing fluids are used in machining operations to cool and lubricate the tool and work-piece. Mist generation is a concern to OSHA, which has recently lowered substantially the workplace standard. The article "Variables Affecting Mist Generation from Metal Removal Fluids" (Lubrication Engr., 2002: 10-17) gave the accompanying data on x = fluid flow velocity for a 5% soluble oil (cm/sec) and y = the extent of mist droplets having diameters smaller than some value:
x: 89 177 189 354 362 442 965
y: .40 .60 .48 .66 .61 .69 .99
a. Make a scatterplot of the data. By R.
b. What is the point estimate of the beta coefficient? (By R.) Interpret it.
c. What is s_e? (By R) Interpret it.
d. Estimate the true average change in mist associated with a 1 cm/sec increase in velocity, and do so in a way that conveys information about precision and reliability.
e. Suppose the fluid velocity is 250 cm/sec. Find the mean of the corresponding y in a way that conveys information about precision and reliability. Use 95% confidence level. Interpret the resulting interval. By hand, as in part d.
f. Suppose the fluid velocity for a specific fluid is 250 cm/sec. Predict the y for that specific fluid in a way that conveys information about precision and reliability. Use 95% prediction level. Interpret the resulting interval. By hand, as in part d.
Answer:
Step-by-step explanation:
a) image attached
b) Lets do the analysis in R , the complete R snippet is as follows
x<- c(89,177,189,354,362,442,965)
y<- c(.4,.6,.48,.66,.61,.69,.99)
# scatterplot
plot(x,y, col="red",pch=16)
# model
fit <- lm(y~x)
summary(fit)
#equation is
#y = 0.4041 + 0.0006211*X
# beta coeffiecients are
fit$coefficients
coef(summary(fit))[, "Std. Error"]
# confidence interval of slope
confint(fit, 'x', level=0.95)
The results are
> summary(fit)
Call:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6 7
-0.05940 0.08595 -0.04151 0.03602 -0.01895 0.01136 -0.01346
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.041e-01 3.459e-02 11.684 8.07e-05 ***
x 6.211e-04 7.579e-05 8.195 0.00044 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.05405 on 5 degrees of freedom
Multiple R-squared: 0.9307, Adjusted R-squared: 0.9168 # model is able to capture 93% of the variation of the data
F-statistic: 67.15 on 1 and 5 DF, p-value: 0.0004403 , p value is less than 0.05 , hence model as a whole is significant
> fit$coefficients
(Intercept) x
0.4041237853 0.0006210758
> coef(summary(fit))[, "Std. Error"]
(Intercept) x
3.458905e-02 7.579156e-05
> confint(fit, 'x', level=0.95)
2.5 % 97.5 %
x 0.0004262474 0.0008159042
c)
> x=c(89,177,189,354,362,442,965)
> y=c(0.40,0.60,0.48,0.66,0.61,0.69,0.99)
>
> ### linear model
> model=lm(y~x)
> summary(model)
Call:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6 7
-0.05940 0.08595 -0.04151 0.03602 -0.01895 0.01136 -0.01346
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.041e-01 3.459e-02 11.684 8.07e-05 ***
x 6.211e-04 7.579e-05 8.195 0.00044 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.05405 on 5 degrees of freedom
Multiple R-squared: 0.9307, Adjusted R-squared: 0.9168
F-statistic: 67.15 on 1 and 5 DF, p-value: 0.0004403
s_e is the Residual standard error from the model and its estimated value is 0.05405. s_e is the standard deviation of the model.
d) 95% confidence interval
> confint(model, confidence=0.95)
2.5 % 97.5 %
(Intercept) 0.3152097913 0.4930377793
x 0.0004262474 0.0008159042
Comment: The estimated confidence interval of slope of x does not include zero. Hence, x has the significant effect on y at 0.05 level of significance.
e)
> predict(model, newdata=data.frame(x=250), interval="confidence", level=0.95)
fit lwr upr
1 0.5593927 0.5020485 0.616737
f)
> predict.lm(model, newdata=data.frame(x=250), interval="prediction", level=0.95)
fit lwr upr
1 0.5593927 0.4090954 0.7096901
verify clairaut's theorem for u=e^xysiny
A professor gives her 100 students an exam; scores are normally distributed. The section has an average exam score of 80 with a standard deviation of 6.5. What percentage of the class has an exam score of A- or higher (defined as at least 90)? Type your calculations along with your answer for full credit; round your final percentage to two decimal places.
Answer:
6.18% of the class has an exam score of A- or higher.
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 = 80, \sigma = 6.5[/tex]
What percentage of the class has an exam score of A- or higher (defined as at least 90)?
This is 1 subtracted by the pvalue of Z when X = 90. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{90 - 80}{6.5}[/tex]
[tex]Z = 1.54[/tex]
[tex]Z = 1.54[/tex] has a pvalue of 0.9382
1 - 0.9382 = 0.0618
6.18% of the class has an exam score of A- or higher.
PLEASE HELP MEEEEEE!!!!!
Answer:
Read below
Step-by-step explanation:
To graph the inequality, place an open circle on -2.5 because there is no line under the > sign. Draw the arrow pointing to the right because the inequality reads "x is greater than -2.5.
As for the check box questions, only B and C should be checked. The arrow points right, and the circle is open.