Lincoln is measuring the angles of quadrilateral WXYZ to determine whether it is congruent to the quadrilateral below.

Mark has a 25 ounce drink. He drinks 5 ounces. What percent of the drink does he have left?

valina [46]

Answer: $80\%$

Step-by-step explanation:

This problem can be solved by the Rule of Three, which is a mathematical rule to find out an amount that is with another quantity given in the same relation as other two also known.

In this case, the 25 ounce drink represents the $100 \%$ . So, if Mark drinks 5 ounces, this means he has 20 ounces left and we have to find the percent this represents:

$25 oz$ ------- $100\%$

$20 oz$ -------- $?$

$?=\frac{(20 oz)(100\%)}{25 oz}$

$?=80\%$ This is the percent of the drink Mark has left

4 0

2 years ago

Consider the initial value problem: 2ty′=8y, y(−1)=1. Find the value of the constant C and the exponent r so that y=Ctr is the s

VikaD [51]

The correct question is:

Consider the initial value problem

2ty' = 8y, y(-1) = 1

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 8y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 8y = 0

Implies

2td(Ct^r)/dt - 8(Ct^r) = 0

2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 8 = 0 or r = 8/2 = 4

Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

C = 1

So, y = t^4

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

and ∂F/∂y are continuous functions defined on a region R. Then there exists a number δ2

(possibly smaller than δ1) so that the solution y = f(x) to (1) is

the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

c) Firstly, we write the differential equation 2ty' = 8y in standard form as

y' - (4/t)y = 0

0 is always continuous, but -4/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

7 0

2 years ago

A principal of $2000 is placed in a savings account at 3% per annum compounded annually. How much is in the account after one ye

natita [175]

Answer:

Step-by-step explanation:

After one year

A=p(1+r/n)^nt

=2000(1+0.03/12)^12*1

=2000(1+0.0025)^12

=2000(1.0025)^12

=2000(1.0304)

=$2060.8

After two-years

A=p(1+r/n)^nt

=2060.8(1+0.03/12)^12*2

=2060.8(1+0.0025)^24

=2060.8(1.0025)^24

=2060.8(1.0618)

=$2188.157

After three years

A=p(1+r/n)^nt

=2188.157(1+0.03/12)^12*3

=2188.157(1+0.0025)^36

=2188.157(1.0025)^36

=2188.157(1.0941)

=$2394.063

8 0

2 years ago

A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the

lbvjy [14]

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

Step-by-step explanation:

(a)

Let X = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

X = 610

n = 1000

Confidence level = 90%

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{610}{1000}=0.61$

The critical value of z for a 90% confidence level is:

$z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645$

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)$

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

Interpretation:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let X = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

X = 440

n = 1000

Confidence level = 95%

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{440}{1000}=0.44$

The critical value of z for a 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)$

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend is (0.42, 0.46).

Interpretation:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

The confidence level
Sample size
Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of z in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

4 0

2 years ago

Box A is 567 inches high, 678 inches wide, and 789 inches long. It has a maximum capacity of 1,200 marbles, Box B has three time

kotykmax [81]

Answer:

1701 high

2712 wide

789 long

= 4413

Step-by-step explanation:

3 0

2 years ago