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vivado [14]
2 years ago
11

How do u make 25 m to *what* cm?

Mathematics
2 answers:
vodka [1.7K]2 years ago
7 0

Answer:

2500 cm

Step-by-step explanation:

1 meter = 100 cm

25m x 100 = 2500 cm

Lera25 [3.4K]2 years ago
6 0

Answer:

25cm is equal to 2500 cm because 1 m is 100 cm

Step-by-step explanation:

Please, may i have brainly? i need 4 more!

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Select the expression that has a value of 4
Harrizon [31]

Answer:

The answer is letter D.

Step-by-step explanation:

320/80

=4

4 0
2 years ago
Factor the expression completely 8x^2-18 Please help
nirvana33 [79]

Answer:

2 (2x - 3) x (2x + 3)

4 0
3 years ago
Direct variation or not a direct variation<br><br> y=5/x
MrRissso [65]

Answer:

not direct variation

Step-by-step explanation:

The equation for x and y in direct variation is

y = kx ← k is the constant of variation

y = \frac{5}{x} is not in this form, thus not direct variation.

The equation for x and y varying inversely is

y = \frac{k}{x} ← k is the constant of variation

y = \frac{5}{x} is in this form and represents inverse variation

7 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
In Mr. Siegel’s class, 15% of the students are in the Scrapbooking Club and 45% are in the Cooking Club. His remaining students
Oksanka [162]

Answer:

Step-by-step explanation:

15% = 15/100= 0.15

45% = 45/100 = o.45

Remaining students: 40% = 40/100 = 0.4%

6 0
2 years ago
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