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2 years ago

Is the answer to -8.8-3.4 positive or negative? I know. It is 12.2 but I don't know if it's -12.2 or +12.2

1 answer:

6 0

It is negative.

When you subtract a negative a positive number from a negative number, the result is positive. The equation can be rewritten as -8.8-(+3.4).

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the legs of a right triangle are 10 centimeters and 24 centimeters long. what is the length of the hypotenuse?

igomit [66]

Pytharorean therum...can only be used on right triangles
a^2 + b^2 = c^2...a and b r ur legs and c is ur hypotenuse
10^2 + 24^2 = c^2
100 + 576 = c^2
676 = c^2
sqrt 676 = c
26 = c <== ur hypotenuse is 26 cm

6 0

2 years ago

Read 2 more answers

A square has a perimeter of 12x + 52 units . Which expression represents the side length of the square in units?

Lapatulllka [165]

Answer:

3x+13

Step-by-step explanation:

Perimeter of square = 4 * length

Length = Perimeter/4

Length = (12x + 52)/4

Length = 3x+13

5 0

3 years ago

Suppose that a number written as a decimal has an infinite number of no -repeating digits after the decimal point. What is this

bekas [8.4K]

Answer:

irrational

are the numbers like 7.55782902..where the numbers aren't repeated

3 0

3 years ago

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total s

garri49 [273]

Answer

a. The expected total service time for customers = 70 minutes

b. The variance for the total service time = 700 minutes

c. It is not likely that the total service time will exceed 2.5 hours

Step-by-step explanation:

This question is incomplete. I will give the complete version below and proceed with my solution.

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) Is it likely that the total service time will exceed 2.5 hours?

Reference

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour.

From the information supplied, we denote that

$X=$ Customers that arrive within the hour

and since $X$ follows a Poisson distribution with mean $\alpha$ = 7

Therefore,

$E(X)= 7$

$& V(X)=7$

Let $Y =$ the total service time for customers arriving during the 1 hour period.

Now, since it takes approximately ten minutes to serve each customer,

$Y=10X$

For a random variable $X$ and a constant $c$ ,

$E(cX)=cE(X)\\V(cX)=c^2V(X)$

Thus,

$E(Y)=E(10X)=10E(X)=10*7=70\\V(Y)=V(10X)=100V(X)=100*7=700$

Therefore the expected total service time for customers = 70 minutes

and the variance for serving time = 700 minutes

Also, the probability of the distribution $Y$ is,

$p_Y(y)=p_x(\frac{y}{10} )\frac{dx}{dy} =\frac{\alpha^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-\alpha } \frac{1}{10}\\ =\frac{7^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-7 } \frac{1}{10}$

So the probability that the total service time exceeds 2.5 hrs or 150 minutes is,

$P(Y>150)=\sum^{\infty}_{k=150} {p_Y} (k) =\sum^{\infty}_{k=150} \frac{7^{\frac{k}{10} }}{(\frac{k}{10})! }.e^{-7} .\frac{1}{10} \\=\frac{7^{\frac{150}{10} }}{(\frac{150}{10})! } .e^{-7}.\frac{1}{10} =0.002$

0.002 is small enough, and the function $\frac{7^{\frac{k}{10} }}{(\frac{k}{10} )!} .e^{-7}.\frac{1}{10}$ gets even smaller when $k$ increases. Hence the probability that the total service time exceeds 2.5 hours is not likely to happen.

3 0

2 years ago

Members of the track team can run 400 min an average time of 64.6 seconds. The

vova2212 [387]

Answer:

T maximum=T average -7.8 seconds

T minimum=T average +7.8 seconds

Step-by-step explanation:

Calculation for the equation that can be

use to find the maximum and minimum times for the track team

Using this equation to find the maximum times for the track team

T maximum=T average -7.8 seconds

T maximum=64.6 seconds-7.8 seconds

Using this equation to find the minimum times for the the track team

T minimum=T average +7.8 seconds

T minimum=64.6 seconds +7.8 seconds

Therefore the equation for the maximum and minimum times for the track team are :

T maximum=T average -7.8 seconds

T minimum=T average +7.8 seconds

6 0

2 years ago