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3 years ago

Solve plz be quick

Mathematics

2 answers:

ivolga24 [154]3 years ago

8 0

72
54
108
2
(Equations used to solve are listed in photo)

Ratling [72]3 years ago

3 0

Answer:

9(10-2)=

90-18=72

11×7-23=54

100-15+23=108

16/8=2

Step-by-step explanation:

prove me wrong

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Help me please 2-4x=x+16

Elina [12.6K]

2-4x=x+16
-5x=13

x=-2.6

5 0

3 years ago

what is the rate of change and initial value for this graph?

MAXImum [283]

Answer:

4/7

Step-by-step explanation:

the rate of change (aka slope) is the change in y over change in x and as y goes up 4 x goes up 7 so the rate of change is 4/7

have a great day :D

8 0

3 years ago

If Anna walled for 5 miles and then turned around and went back what is her displacement?

damaskus [11]

If she went back 5 miles than the displacement would be 0.

Just subtract the amount of miles she went back from the miles she went forward 5 - 5 = 0

3 0

3 years ago

Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1pm and 7p

Ymorist [56]

Answer:

$E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours$

$Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2$

Step-by-step explanation:

Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:

$A\sim Uniform(1 ,7)$

And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:

$B\sim Exp(\lambda=\frac{1}{30 min})$

Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.

So we are interested on the expected value of A+B, like this

$E(A +B)$

Since the two random variables are assumed independent, then we have this

$E(A+B) = E(A)+E(B)$

So we can find the individual expected values for each distribution and then we can add it.

For ths uniform distribution the expected value is given by $E(X) =\frac{a+b}{2}$ where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

$E(A)=\frac{1+7}{2}=4 hours$

The expected value for the exponential distirbution is given by :

$E(X)= \int_{0}^\infty x \lambda e^{-\lambda x} dx$

If we use the substitution $y=\lambda x$ we have this:

$E(X)=\frac{1}{\lambda} \int_{0}^\infty y e^{-\lambda y} dy =\frac{1}{\lambda}$

Where X represent the random variable and $\lambda$ the parameter. If we apply this formula to our case we got:

$E(B) =\frac{1}{\lambda}=\frac{1}{\frac{1}{30}}=30min$

We can convert this into hours and we got E(B) =0.5 hours, and then we can find:

$E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours$

And in order to find the variance for the random variable A+B we can find the individual variances:

$Var(A)= \frac{(b-a)^2}{12}=\frac{(7-1)^2}{12}=3 hours^2$

$Var(B) =\frac{1}{\lambda^2}=\frac{1}{(\frac{1}{30})^2}=900 min^2 x\frac{1hr^2}{3600 min^2}=0.25 hours^2$

We have the following property:

$Var(X+Y)= Var(X)+Var(Y) +2 Cov(X,Y)$

Since we have independnet variable the Cov(A,B)=0, so then:

$Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2$

3 0

3 years ago

If Samantha can pay off her loan in 36 months at a 10% interest rate rather than in 48 months at a 12% interest rate, how much m

Ymorist [56]

Hi there
The simple interest formula is
I=prt
I interest changes
P amount of the loan 6000
R interest rate
T time( number of months/12 months)

The interest in 36 months at a 10%
I=6,000×0.1×(36÷12)=1,800
The interest in 48 months at a 12%
6,000×0.12×(48÷12)=2,880
she will save
2,880−1,800=1,080

Good luck!

7 0

3 years ago

Solve *plz be quick*

Solve plz be quick