How do u solve this equation help me

Answers

Answer 1

Answer: In this equation P=1

Answer 2

Answer: The first answer is right, but the working goes like this:
-p - 10 = -11
+ p
-10 = -11 + p
+ 11
1 = p

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Solve the system of equations below by graphing them with a pencil andpaper. Enter your answer as an ordered pair.
y=-x+ 2
y=-2x+6

Answers

Answer:

x=2 y=0

Step-by-step explanation:

Explanation:

just put y= -x in second equation x comes out to be 2. Then put x=2 y comes out to be 0

Consider a home mortgage of $250,000 at a fixed APR of 4.5% for 25 years.a. Calculate the monthly payment.
b. Determine the total amount paid over the term of the loan.
c. Of the total amount paid, what percentage is paid toward the principal and what
percentage is paid for interest.
a. The monthly payment is $ 1389.58.
(Do not round until the final answer. Then round to the nearest cent as needed.)
b. The total amount paid over the term of the loan is $
(Round to the nearest cent as needed.)
ANSWER ALL PARTS PLEASE (A,B,C)

Answers

ALL THE ANSWERS AND EXPLANATION ARE GIVEN BY CHATGPT LOOK IN THE FILE ATTACHED

g let X be a normally distributed random variable with mean 3 and variance 4. a) Let Y = 5X+2. What is the distribution of Y? What are its mean and variance? b) Find P(Y<10). Find P(X<10). c) What is the 99th percentile of the distribution of Y? d) What is the 99th percentile of the distribution of X? e) What is the distribution of W = exp(Y)? What are its mean and variance?

Answers

a. Let $F_X(x)$ be the CDF of $X$ . The CDF of $Y$ is

$F_Y(y)=P(Y\le y)=P(5X+2\le y)=P\left(X\le\frac{y-2}5\right)=F_X\left(\frac{y-2}5\right)$

which is to say, $Y$ is also normally distributed, but with different parameters. In particular,

$E[Y]=E[5X+2]=5E[X]+2=17$

$\mathrm{Var}[Y]=\mathrm{Var}[5X+2]=5^2\mathrm{Var}[X]=100$

b. Using the appropriate CDFs, we have

$P(Y<10)=F_Y(10)=F_X\left(\frac{10-2}5\right)=F_X(1.6)\approx0.242$

$P(X<10)=F_X(10)\approx0.9998$

c. The 99th percentile for any distribution $D$ is the value of $d_(0.99)$ such that $P(D\le d_(0.99))=0.99$ , i.e. all values of $d$ below $d_(0.99)$ make up the lower 99% of the distribution.

We have

$P(Y\le y_(0.99))=0.99\implies y_(0.99)\approx40.26$

d. On the other hand, the 99th percentile for $X$ is

$P(X\le x_(0.99))=0.99\implies x_(0.99)\approx7.653$

e. We have

$F_W(w)=P(W\le w)=P\left(e^Y\le w\right)=P(Y\le\ln w)=F_Y(\ln w)$

which suggests that $\ln W$ is normally distributed, or $W$ is log-normally distributed. Recall that the moment-generating function for $Y$ is

$M_Y(t)=\exp\left(17t+\frac{100t^2}2\right)$

But we also have

$M_Y(t)=E[e^(tY)]=E[e^(t\ln W)]=E[W^t]$

Then

$E[W]=M_Y(1)=e^(67)$

and

$E[W^2]=M_Y(2)=e^(234)\implies\mathrm{Var}[W]=E[W^2]-E[W]^2=e^(234)-e^(134)$

Can you please help me ASAP. its due very soon.

Answers

The final answer would be 3.5 inches tall or if you need fractions it would be 3 and 1/2. Here’s the explanation: the candle is originally eight inches tall and for every hour it burns it loses 3/4 of an inch which is .75. It will be burning till 11:00 at night which means it will be burning for six hours. All you need to do is multiply how much of the candle is burned every hour (.75) by how many hours. This will give you 4.5 inches burned by 11:00. But the question asked for how tall the candle will be, so you simply just subtract how much of the candle will have burned(4.5 inches) by the height (8 inches) and you get 3.5 inches.

Answer:

4.5 inches

Step-by-step explanation:

8-3/4-3/4-3/4-3/4-3/4-3/4= 4 2/4

When using 3 letters to make an angle the middle letter is