Answer:
[tex]\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
The expected values for all the categories is :
[tex] E_i =\frac{150}{3}=50[/tex]
And then the statistic would be given by:
[tex]\chi^2 = \frac{(40-50)^2}{50}+\frac{(60-50)^2}{50}+\frac{(50-50)^2}{50}=4[/tex]
And the best option would be:
b. 4
Step-by-step explanation:
For this problem we have the following observed values:
Yes 40 No 60 No Opinion 50
And we want to test the following hypothesis:
Null hypothesis: All the opinions are uniformly distributed
Alternative hypothesis: Not All the opinions are uniformly distributed
And for this case the statistic would be given by:
[tex]\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
The expected values for all the categories is :
[tex] E_i =\frac{150}{3}=50[/tex]
And then the statistic would be given by:
[tex]\chi^2 = \frac{(40-50)^2}{50}+\frac{(60-50)^2}{50}+\frac{(50-50)^2}{50}=4[/tex]
And the best option would be:
b. 4
Perform the indicated operation.
Answer:
√75 = 5√3 and √12 = 2√3 so √75 + √12 = 5√3 + 2√3 = 7√3.
Answer:
[tex] 7\sqrt{3} [/tex]
Step-by-step explanation:
[tex] \sqrt{12} \: can \: be \: simplified \: as \: 2 \sqrt{3} \: and \: \sqrt{75} \: canbe \: simplified \: as \: 5 \sqrt{3} \\ after \: simplifying \: we \: can \: add \: them \: up \\ 2 \sqrt{3} + 5 \sqrt{3} = 7 \sqrt{3} [/tex]
The Ericsson method is one of several methods claimed to increase the likelihood of a baby girl. In a clinical trial, results could be analyzed with a formal hypothesis test with the alternative hypothesis of pgreater than0.5,which corresponds to the claim that the method increases the likelihood of having a girl, so that the proportion of girls is greater than 0.5. If you have an interest in establishing the success of the method, which of the following P-values would you prefer: 0.999, 0.5, 0.95, 0.05, 0.01, 0.001? Why?
Answer:
0.001
Step-by-step explanation:
Here, the aim is to support the null hypothesis, Ha. Where Ha: p > 0.5. Which means we are to reject null hypothesis H0. Where H0: p = 0.5.
The higher the pvalue, the higher the evidence of success. We know If the pvalue is less than level of significance, the null hypothesis H0 is rejected.
Hence the smallest possible value 0.001 is preferred as the pvalue because it corresponds to the sample evidence that most strongly supports the alternative hypothesis that the method is effective
Buchtal, a manufacturer of ceramic tiles, reports on average 2.3 job-related accidents per year. Accident categories include trip, fall, struck by equipment, transportation, and handling. The number of accidents is approximately Poisson. Please upload your work for all of the parts at the end. a) What is the probability that more than one accident occurs per year? Include at least 3 decimal places in your answer. Submit Answer Tries 0/5 b) Suppose that 5 years are randomly selected. What is the expected number of accidents in this time period? Submit Answer Tries 0/5 c) What is the standard deviation of the number of accidents in 5 years? Submit Answer Tries 0/5 d) What is the probability that exactly 8 accidents occur in 5 years? Include at least 3 decimal places in your answer. If you get an error on your calculator, please use an online source like Wolfram Alpha to calculate the number. Submit Answer Tries 0/5
Answereippcb.jrc.ec.europa.eu
Step-by-step explanation:
this I the wed go on it and you will get your answer
CAN SOMEONE HELP ME ASAP
A. 5
B. 53‾√53
C. 10
D. 103√3
Answer:
n = 5
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
tan theta = opp/ adj
tan 30 = n/ 5 sqrt(3)
5 sqrt(3) tan 30 = n
5 sqrt(3) * 1/ sqrt(3) = n
5 = n
A 12 sided die is rolled the set of equally likely outcomes is 123 456-789-10 11 and 12 find the probability of rolling a number greater than three
Answer:
6
Step-by-step explanation:
nerd physics
You want to install a 1 1 yd wide walk around a circular swimming pool. The diameter of the pool is 23 yd. What is the area of the walk? Use 3.14 for pi π.
Complete Question:
You want to install a 1 yd wide walk around a circular swimming pool. The diameter of the pool is 23 yd. What is the area of the walk? Use 3.14 for pi π.
Answer:
75.36 square yard
Step-by-step explanation:
From the question,
The diameter of this circular pool inside is 23 yd.
This means that the radius = Diameter/2 = 23yd/2 = 11.5 yd.
The formula for the area of a circle =
A = πr²
A = π(11.5)²
A =3.14 × 11.5²
A = 415.265 yd²
This is the Area of the inner circle.
We were told in the question also that he wants to install a walk of 1 yard
Hence, the radius of outer circle =
radius of inner circle +length of the walk
11.5yard + 1 yard
= 12.5 yard
A = πr²
A = 3.14 × (12.5)²
A = 490.625yd²
Area of the walk = Area of the Outer circle - Area of the inner circle
= (490.625 - 415.265)yd = 75.36 yd²
Therefore, the area of the walk is 75.36 square yards.
The Ball Corporation's beverage can manufacturing plant in Fort Atkinson, Wisconsin, uses a metal supplier that provides metal with a known thickness standard deviation σ = .000586 mm. Assume a random sample of 59 sheets of metal resulted in an x¯ = .2905 mm. Calculate the 95 percent confidence interval for the true mean metal thickness.
Answer:
The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.96\frac{0.000586}{\sqrt{59}} = 0.0002[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 0.2905 - 0.0002 = 0.2903 mm
The upper end of the interval is the sample mean added to M. So it is 0.2905 + 0.0002 = 0.2907 mm
The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm
Suppose that the relationship between the tax rate t on imported shoes and the total sales S (in millions of dollars) is given by the function below. Find the tax rate t that maximizes revenue for the government. (Round your answer to three decimal places.)
S(t) = 7 â 6(cubedroot(t))
Answer:
66.992%
Step-by-step explanation:
[tex]Sales, S(t)=7-6\sqrt[3]{t}[/tex]
Since we want to maximize revenue for the government
Government's Revenue= Sales X Tax Rate
[tex]R(t)=t \cdot S(t)\\R(t)=t(7-6\sqrt[3]{t})\\=7t-6t^{1+1/3}\\R(t)=7t-6t^{4/3}[/tex]
To maximize revenue, we differentiate R(t) and equate it to zero to solve for its critical points. Then we test that this critical point is a relative maximum for R(t) using the second derivative test.
Now:
[tex]R'(t)=7-6*\frac{4}{3} t^{4/3-1}\\=7-8t^{1/3}[/tex]
Setting the derivative equal to zero
[tex]7-8t^{1/3}=0\\7=8t^{1/3}\\t^{1/3}=\dfrac{7}{8} \\t=(\frac{7}{8})^3\\t=0.66992[/tex]
Next, we determine that t=0.6692 is a relative maximum for R(t) using the second derivative test.
[tex]R''(t)=-8*\frac{1}{3} t^{1/3-1}\\R''(t)=-\frac{8}{3} t^{-2/3}[/tex]
R''(0.6692)=-3.48 (which is negative)
Therefore, t=0.66992 is a relative maximum for R(t).
The tax rate, t that maximizes revenue for the government is:
=0.66992 X 100
t=66.992% (correct to 3 decimal places)
We are standing on the top of a 320 foot tall building and launch a small object upward. The object's vertical altitude, measured in feet, after t seconds is h ( t ) = − 16 t 2 + 128 t + 320 . What is the highest altitude that the object reaches?
Answer:
The highest altitude that the object reaches is 576 feet.
Step-by-step explanation:
The maximum altitude reached by the object can be found by using the first and second derivatives of the given function. (First and Second Derivative Tests). Let be [tex]h(t) = -16\cdot t^{2} + 128\cdot t + 320[/tex], the first and second derivatives are, respectively:
First Derivative
[tex]h'(t) = -32\cdot t +128[/tex]
Second Derivative
[tex]h''(t) = -32[/tex]
Then, the First and Second Derivative Test can be performed as follows. Let equalize the first derivative to zero and solve the resultant expression:
[tex]-32\cdot t +128 = 0[/tex]
[tex]t = \frac{128}{32}\,s[/tex]
[tex]t = 4\,s[/tex] (Critical value)
The second derivative of the second-order polynomial presented above is a constant function and a negative number, which means that critical values leads to an absolute maximum, that is, the highest altitude reached by the object. Then, let is evaluate the function at the critical value:
[tex]h(4\,s) = -16\cdot (4\,s)^{2}+128\cdot (4\,s) +320[/tex]
[tex]h(4\,s) = 576\,ft[/tex]
The highest altitude that the object reaches is 576 feet.
The curvature of a plane parametric curve x = f(t), y = g(t) is $ \kappa = \dfrac{|\dot{x} \ddot{y} - \dot{y} \ddot{x}|}{[\dot{x}^2 + \dot{y}^2]^{3/2}}$ where the dots indicate derivatives with respect to t. Use the above formula to find the curvature. x = 6et cos(t), y = 6et sin(t)
Answer:
The curvature is modelled by [tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex].
Step-by-step explanation:
The equation of the curvature is:
[tex]\kappa = \frac{|\dot {x}\cdot \ddot {y}-\dot{y}\cdot \ddot{x}|}{[\dot{x}^{2}+\dot{y}^{2}]^{\frac{3}{2} }}[/tex]
The parametric componentes of the curve are:
[tex]x = 6\cdot e^{t} \cdot \cos t[/tex] and [tex]y = 6\cdot e^{t}\cdot \sin t[/tex]
The first and second derivative associated to each component are determined by differentiation rules:
First derivative
[tex]\dot{x} = 6\cdot e^{t}\cdot \cos t - 6\cdot e^{t}\cdot \sin t[/tex] and [tex]\dot {y} = 6\cdot e^{t}\cdot \sin t + 6\cdot e^{t} \cdot \cos t[/tex]
[tex]\dot x = 6\cdot e^{t} \cdot (\cos t - \sin t)[/tex] and [tex]\dot {y} = 6\cdot e^{t}\cdot (\sin t + \cos t)[/tex]
Second derivative
[tex]\ddot{x} = 6\cdot e^{t}\cdot (\cos t-\sin t)+6\cdot e^{t} \cdot (-\sin t -\cos t)[/tex]
[tex]\ddot x = -12\cdot e^{t}\cdot \sin t[/tex]
[tex]\ddot {y} = 6\cdot e^{t}\cdot (\sin t + \cos t) + 6\cdot e^{t}\cdot (\cos t - \sin t)[/tex]
[tex]\ddot{y} = 12\cdot e^{t}\cdot \cos t[/tex]
Now, each term is replaced in the the curvature equation:
[tex]\kappa = \frac{|6\cdot e^{t}\cdot (\cos t - \sin t)\cdot 12\cdot e^{t}\cdot \cos t-6\cdot e^{t}\cdot (\sin t + \cos t)\cdot (-12\cdot e^{t}\cdot \sin t)|}{\left\{\left[6\cdot e^{t}\cdot (\cos t - \sin t)\right]^{2}+\right[6\cdot e^{t}\cdot (\sin t + \cos t)\left]^{2}\right\}^{\frac{3}{2}}} }[/tex]
And the resulting expression is simplified by algebraic and trigonometric means:
[tex]\kappa = \frac{72\cdot e^{2\cdot t}\cdot \cos^{2}t-72\cdot e^{2\cdot t}\cdot \sin t\cdot \cos t + 72\cdot e^{2\cdot t}\cdot \sin^{2}t+72\cdot e^{2\cdot t}\cdot \sin t \cdot \cos t}{[36\cdot e^{2\cdot t}\cdot (\cos^{2}t -2\cdot \cos t \cdot \sin t +\sin^{2}t)+36\cdot e^{2\cdot t}\cdot (\sin^{2}t+2\cdot \cos t \cdot \sin t +\cos^{2} t)]^{\frac{3}{2} }}[/tex]
[tex]\kappa = \frac{72\cdot e^{2\cdot t}}{[72\cdot e^{2\cdot t}]^{\frac{3}{2} } }[/tex]
[tex]\kappa = [72\cdot e^{2\cdot t}]^{-\frac{1}{2} }[/tex]
[tex]\kappa = 72^{-\frac{1}{2} }\cdot e^{-t}[/tex]
[tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex]
The curvature is modelled by [tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex].
Assume that the random variable X is normally distributed, with mean 60 and standard deviation 16. Compute the probability P(X < 80). Group of answer choices
Answer:
P(X < 80) = 0.89435.
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 = 60, \sigma = 16[/tex]
P(X < 80)
This is the pvalue of Z when X = 80. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 60}{16}[/tex]
[tex]Z = 1.25[/tex]
[tex]Z = 1.25[/tex] has a pvalue of 0.89435.
So
P(X < 80) = 0.89435.
A large mixing tank initially contains 1000 gallons of water in which 30 pounds of salt have been dissolved. Another brine solution is pumped into the tank at the rate of 4 gallons per minute, and the resulting mixture is pumped out at the same rate. The concentration of the incoming brine solution is 2 pounds of salt per gallon. If represents the amount of salt in the tank at time t, the correct differential equation for A is:__________.A.) dA/dt = 4 - .08AB.) dA/dt = 8 -.04AC.) dA/dt = 4-.04AD.) dA/dt = 2-.04AE.) dA/dt = 8-.02A
Answer:
(B)[tex]\dfrac{dA}{dt}=8-0.004A[/tex]
Step-by-step explanation:
Volume of fluid in the tank =1000 gallons
Initial Amount of Salt in the tank, A(0)= 30 pounds
Incoming brine solution of concentration 2 pounds of salt per gallon is pumped in at a rate of 4 gallons per minute.
Rate In=(concentration of salt in inflow)(input rate of brine)
[tex]=(2\frac{lbs}{gal})( 4\frac{gal}{min})=8\frac{lbs}{min}[/tex]
The resulting mixture is pumped out at the same rate, therefore:
Rate Out =(concentration of salt in outflow)(output rate of brine)
[tex]=(\frac{A(t)}{1000})( 4\frac{gal}{min})=\frac{A}{250}[/tex]
Therefore:
The rate of change of amount of salt in the tank,
[tex]\dfrac{dA}{dt}=$Rate In-Rate out\\\dfrac{dA}{dt}=8-\dfrac{A}{250}\\\dfrac{dA}{dt}=8-0.004A[/tex]
A toy falls from a window 80 feet above the ground. How long does it take the toy to hit the ground?
Answer:
2.24 s
Step-by-step explanation:
Given:
Δy = 80 ft
v₀ = 0 ft/s
a = 32 ft/s²
Find: t
Δy = v₀ t + ½ at²
80 ft = (0 ft/s) t + ½ (32 ft/s²) t²
t = 2.24 s
Please answer this correctly
Answer:
2/3
Step-by-step explanation:
Total sides = 6
Number 5 and all even numbers = 1+3
=> 4
P(5 or even ) = 4/6
=> 2/3
Explain in your own words why a polynomial can’t be a quadratic if a= 0?
If [tex]a = 0[/tex], then [tex]y = ax^2+bx+c[/tex] turns into [tex]y = 0x^2+bx+c[/tex]. That [tex]0x^2[/tex] term goes away because it turns into 0, and adding 0 onto anything does not change the expression.
So [tex]y = 0x^2+bx+c[/tex] turns into [tex]y = bx+c[/tex] which is a linear equation (b is the slope, c is the y intercept). It is no longer a quadratic as quadratic equations always graph out a curved parabola.
As an example, you could graph out [tex]y = 0x^2+3x+4[/tex] and note how it's the exact same as [tex]y = 3x+4[/tex], both of which are straight lines through the two points (0,4) and (1,7).
Which best compares the volumes of the two cylinders? Geometry
Answer:
The correct answer would be C
Step-by-step explanation:
please mark brainliest
The choice which best compares the volume of the cylinders is; Choice B; The volume of cylinder B is the same as that of cylinder A.
Which best compares the volumes of the two cylinders?From geometry, It can be concluded that the volume of a solid shape is the product of its cross sectional area and the height over which the area spans. On this note, since the volume of a cylinder is dependent on the radius and height of the cylinder, both cylinders have equal volumes.
Read more on cylinders;
https://brainly.com/question/9554871
#SPJ2
Add. Answer as a fraction. Do not include spaces in your answer. Do not include spaces in your answer.
Answer: 49/9
Step-by-step explanation: 42/9 + 7/9 = 49/9
Make first fraction into improper fraction with the same common dominator as 7/9 and add them both
Hope this helps:)
Answer:
49/9
Step-by-step explanation:
The mean and standard deviation of a random sample of n measurements are equal to 34.5 and 3.4, respectively.A. Find a 95 % confidence interval for μ if n=49.B. Find a 95% confidence interval for μ if n=196.C. Find the widths of the confidence intervals found in parts a and b.D. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient fixed?1. Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 4.2. Quadrupling the sample size while holding the confidence coefficient fixed increases the width of the confidence interval by a factor of 2.3. Quadrupling the sample size while holding the confidence coefficient fixed increases the width of the on confidence interval by a factor of 4.4. Quadrupling the sample size while holding the confidence coefficient fixed does not affect the width of the confidence interval.5. Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 2.
Answer:
a. The 95% confidence interval for the mean is (33.52, 35.48).
b. The 95% confidence interval for the mean is (34.02, 34.98).
c. n=49 ⇒ Width = 1.95
n=196 ⇒ Width = 0.96
Note: it should be a factor of 2 between the widths, but the different degrees of freedom affects the critical value for each interval, as the sample size is different. It the population standard deviation had been used, the factor would have been exactly 2.
d. 5. Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 2.
Step-by-step explanation:
a. We have to calculate a 95% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=34.5.
The sample size is N=49.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{49}}=\dfrac{3.4}{7}=0.486[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=49-1=48[/tex]
The t-value for a 95% confidence interval and 48 degrees of freedom is t=2.011.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=2.011 \cdot 0.486=0.98[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = 34.5-0.98=33.52\\\\UL=M+t \cdot s_M = 34.5+0.98=35.48[/tex]
The 95% confidence interval for the mean is (33.52, 35.48).
b. We have to calculate a 95% confidence interval for the mean.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{196}}=\dfrac{3.4}{14}=0.243[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=196-1=195[/tex]
The t-value for a 95% confidence interval and 195 degrees of freedom is t=1.972.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=1.972 \cdot 0.243=0.48[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = 34.5-0.48=34.02\\\\UL=M+t \cdot s_M = 34.5+0.48=34.98[/tex]
The 95% confidence interval for the mean is (34.02, 34.98).
c. The width of the intervals is:
[tex]n=49\rightarrow UL-LL=33.52-35.48=1.95\\\\n=196\rightarrow UL-LL=34.02-34.98=0.96[/tex]
d. The width of the intervals is decreased by a factor of √4=2 when the sample size is quadrupled, while the others factors are fixed.
PLEASE ANSWER FAST, THANKS! :)
Answer:
Step-by-step explanation:
k = 3 ; 2k + 2 = 2*3 + 2 = 6 + 2 = 8
k = 4; 2k + 2 = 2*4 + 2 = 8 +2 = 10
k =5; 2k + 2 = 2*5 +2 = 10+2 = 12
k=6; 2k +2 = 2*6 + 2 = 12+2 = 14
k = 7 ; 2k + 2 = 2*7 +2 = 14 +2 = 16
k = 8 ; 2k + 2 = 2*8 + 2 = 16 +2 = 18
∑ (2k + 2) = 8 + 10 + 12 + 14 + 16 + 18 = 78
After the last ice age began, the number of animal species in Australia changed rapidly. The relationship between the elapsed time, t, in years, since the ice age began, and the total number of animal species, S year(t), is modeled by the following function: S year(t)=25,000,000⋅(0.78)t Complete the following sentence about the rate of change in the number of species in decades. Round your answer to two decimal places. Every decade, the number of species decays by a factor of
Answer:
Every decade, the number of species decays by a factor of 0.0834.
Step-by-step explanation:
Let be [tex]S(t) = 25,000,000\cdot 0.78^{t}[/tex], [tex]\forall t \geq 0[/tex]. The decay rate per decay is deducted from the following relation:
[tex]\frac{S(t+10)}{S(t)} = \frac{25,000,000\cdot 0.78^{t+10}}{25,000,000\cdot 0.78^{t}}[/tex]
[tex]\frac{S(t+10)}{S(t)} = 0.78^{t+10-t}[/tex]
[tex]\frac{S(t+10)}{S(t)} = 0.78^{10}[/tex]
[tex]\frac{S(t+10)}{S(t)} = 0.0834[/tex]
Every decade, the number of species decays by a factor of 0.0834.
Answer:
28% subtracted
Step-by-step explanation:
khan
A scooter runs 40 km using 1 litre of petrol tje distance covered by it using 15/4 litres of petrol is
Answer:
150 km
Step-by-step explanation:
1 liter ............ 40 km
15/4 liter .........x km
x = 15/4×40/1 = 600/4 = 150 km
A boy has 27 cubes, each with sides the length of 1cm. He uses these cubes to build one big cube. What is the volume of the big cube?
Answer:54
volume:side*side*side
side:1 cm*1 cm *1 cm
answer=icm
Select the correct answer from each drop-down menu.
The given equation has been solved in the table.
Answer: a) additive inverse (addition)
b) multiplicative inverse (division)
Step-by-step explanation:
Step 2: 6 is being added to both sides
Step 4: (3/4) is being divided from both sides
It is difficult to know what options are provided in the drop-down menu without seeing them. If I was to complete a proof and justify each step, then the following justifications would be used:
Step 2: Addition Property of Equality
Step 4: Division Property of Equality
16. How much money will I need to have at retirement so I can withdraw $60,000 a year for 20 years from an account earning 8% compounded annually? a. How much do you need in your account at the beginning b. How much total money will you pull out of the account? c. How much of that money is interest?
Answer:
starting balance: $636,215.95total withdrawals: $1,200,000interest withdrawn: $563,784.05Step-by-step explanation:
a) If we assume the annual withdrawals are at the beginning of the year, we can use the formula for an annuity due to compute the necessary savings.
The principal P that must be invested at rate r for n annual withdrawals of amount A is ...
P = A(1+r)(1 -(1 +r)^-n)/r
P = $60,000(1.08)(1 -1.08^-20)/0.08 = $636,215.95
__
b) 20 withdrawals of $60,000 each total ...
20×$60,000 = $1,200,000
__
c) The excess over the amount deposited is interest:
$1,200,000 -636,215.95 = $563,784.05
Cheryl bought 3.4 pounds of coffee that cost $6.95 per pound . How many did she spend on coffee
Answer:
23.63
Step-by-step explanation:
multiply the cost by the pounds
Answer:
$23.63
Step-by-step explanation:
3.4 X 6.95 = 23.63
helppppppp pleassssseeeeee
Answer:
First blank is 4, second blank is 0
Step-by-step explanation:
divide it :)
Answer:
Yellow box #1=0
Yellow box #1=4
Step-by-step explanation:
The vector matrix[ 27 ]is dilated by a factor of 1.5 and then reflected across the X axis if the resulting matrix is a B then a equals an VE
Correct question:
The vector matrix [ [tex] \left[\begin{array}{ccc}2\\7\end{array}\right] [/tex] is dilated by a factor of 1.5 and then reflected across the x axis. If the resulting matrix is [a/b] then a=??? and b=???
Answer:
a = 3
b = 10.5
Step-by-step explanation:
Given:
Vector matrix = [tex] \left[\begin{array}{ccc}2\\7\end{array}\right] [/tex]
Dilation factor = 1.5
Since the vector matrix is dilated by 1.5, we have:
[tex] \left[\begin{array}{ccc}1.5 * 2\\1.5 * 7\end{array}\right] [/tex]
= [tex] \left[\begin{array}{ccc}3\\10.5\end{array}\right] [/tex]
Here, we are told the vector is reflected on the x axis.
Therefore,
a = 3
b = 10.5
Answer:
a = 3
b = -10.5
Step-by-step explanation:
got a 100% on PLATO
A boat that can travel 18 mph in still water can travel 21 miles downstream in the same amount of time that it can travel 15 miles upstream. Find the speed (in mph) of the current in the river.
Hey there! I'm happy to help!
We see that if the river isn't moving at all the boat can move at 18 mph (most likely because it has an engine propelling it.)
We want to set up a proportion where our 21 miles downstream time is equal to our 15 miles upstream time so we can find the speed. A proportion is basically showing that two ratios are equal. Since our downstream distance and upstream distance can be done in the same amount of time, we will write it as a proportion.
We want to find the speed of the river. We will use r to represent the speed of the river. When going downstream, the boat will go faster, so it will have a higher mph. So, our speed going down is 18+r. When you are going upstream, it's the opposite, so it will be 18-r.
[tex]\frac{distance}{speed} =\frac{21}{18+r} = \frac{15}{18-r}[/tex]
So, how do we figure out what r is now? Well, one nice thing to know about proportions is that the product of the items diagonal from each other equals the product of the other items. Basically, that means that 15(18+r) is equal to 21(18-r). This is a very nice trick to solve proportions quickly. We see that we have made an equation and now we can solve it!
15(18+r)=21(18-r)
We use the distributive property to undo the parentheses.
270+15r=378-21r
We subtract 270 from both sides.
15r=108-21
We add 21 to both sides.
36r=108
We divide both sides by 36.
r=3
Therefore, the speed of the river is 3 mph.
You also could have noticed that 18mph to 21 mph is +3, and 18mph to 15 mph -3 in -3 mph, so the speed of the river is 3 mph. That would have been a quicker way to solve it XD!
Have a wonderful day!
The straight line L has equation y = 1/2x+7 The straight line M is parallel to L and passes through the point (0, 3). Write down an equation for the line M.
Answer:
y = [tex]\frac{1}{2}[/tex] x + 3
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
y = [tex]\frac{1}{2}[/tex] x + 7 ← is in slope- intercept form
with slope m = [tex]\frac{1}{2}[/tex]
Parallel lines have equal slopes
line M crosses the y- axis at (0, 3) ⇒ c = 3
y = [tex]\frac{1}{2}[/tex] x + 3 ← equation of line M
What is the measure of
Answer:
C. 35
55 degrees + 35 degrees= 90 degrees