A conservative vector field 
has curl 
. In this case,
so the vector field is not conservative.
so the vector field is not conservative.
Diego and Amit were trying to solve the equation: x^2-8x=1x 2 −8x=1x, squared, minus, 8, x, equals, 1 Diego said, "I can solve b
1 answer:
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Answer:
a: 28 < µ < 34
Step-by-step explanation:
We need the mean, var, and standard deviation for the data set. See first attached photo for calculations for these...
We get a mean of 222/7 = 31.7143
and a sample standard deviation of: 4.3079
We can now construct our confidence interval. See the second attached photo for the construction steps.
They want a 90% confidence interval. Our sample size is 7, so since n < 30, we will use a t-score. Look up the value under the 10% area in 2 tails column, and degree of freedom is 6 (degree of freedom is always 1 less than sample size for confidence intervals when n < 30)
The t-value is: 1.943
We rounded down to the nearest person in the interval because we don't want to over estimate. It said 28.55, so more than 28 but not quite 29, so if we use 29 as the lower limit, we could over estimate. It's better to use 28 and underestimate a little when considering customer flow.
Answer:
The amount of salt in the tank when it is full of brine is 393.75 pounds.
Step-by-step explanation:
This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).
The main equation that we’ll be using to model this situation is:
Rate of change of <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> exits the tank
where,
Rate at which Q(t) enters the tank = (flow rate of liquid entering) x
(concentration of substance in liquid entering)
Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x
(concentration of substance in liquid exiting)
Let y<em>(t)</em> be the amount of salt (in pounds) in the tank at time <em>t</em> (in seconds). Then we can represent the situation with the below picture.
Then the differential equation we’re after is
is the volume of brine in the tank at time <em>t. </em>To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,
We can then write the initial value problem:
We have a linear differential equation. A first-order linear differential equation is one that can be put into the form
where <em>P</em> and <em>Q</em> are continuous functions on a given interval.
In our case, we have that
The solution process for a first order linear differential equation is as follows.
Step 1: Find the integrating factor, , using
Step 2: Multiply everything in the differential equation by and verify that the left side becomes the product rule
and write it as such.
Step 3: Integrate both sides.
Step 4: Find the value of the constant and solve for the solution y(t).
Now, the tank is full of brine when:
The amount of salt in the tank when it is full of brine is
Answer:
(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].
(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval
(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.
Step-by-step explanation:
We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.
Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of Americans who decide to not go to college = 48%
n = sample of American adults = 331
p = population proportion of Americans who decide to not go to
college because they cannot afford it
<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>
<em />
<u>So, 90% confidence interval for the population proportion, p is ;</u>
P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level
of significance are -1.645 & 1.645}
P(-1.645 < < 1.645) = 0.90
P( <
<
) = 0.90
P( < p <
) = 0.90
<u>90% confidence interval for p</u> = [ ,
]
= [ ,
]
= [0.4348, 0.5252]
(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].
(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.
3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.
So, the margin of error =
= 54.79
n =
n = 3001.88 ≈ 3002
Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.