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3 years ago

9

Write an expression that has three terms, two different variables, and one constant

1 answer:

morpeh [17]3 years ago

7 0

This should do it: 2x + 3y + 4

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How many servings of 3/4 cups are in 32/2cups

NARA [144]

Answer: 3/64

Step-by-step explanation: I think you ask 3/4 divided by 32/2 if you are the answer is 3/64. How to divided is first simplify the expression, if possible, by canceling the common factors. But if your asking me to do 32/2 divided by 3/4 the answer is 64/3 if you want it to be an improper fraction but if you want it to be a mixed fraction it would be 21 1/3. How to divided is first simplify the expression, if possible, by canceling the common factors.

8 0

2 years ago

Solve the equation 3x-2y =7, x+2y= -3

swat32

Answer:

x=1
y= -2

Step-by-step explanation:

3x-2y =7
x+2y= -3
4x=4
x=4/4
x=1
3x-2y=7
3(1)-2y=7
3-2y=7
-2y=7-3
-2y=4
2y=-4
y=-4/2
y= -2
Verify solution
x+2y= -3
1+2y= -3
2y=-3-1
2y= -4
y= -4/2
y= -2

3 0

1 year ago

Read 2 more answers

The toatal charge on 6 is 48 units . what is the charge on each house

Vesna [10]

The charge on each house is 8 units

3 0

3 years ago

The solid figure at the right is made up of two

natima [27]

7x20x11= 1540
30x4x11=1320
1540+1320=2860

5 0

3 years ago

Read 2 more answers

At an outpatient mental health clinic, appointment cancellations occur at a mean rate of 1.5 per day on a typical Wednesday. Let

Kamila [148]

Answer:

a) Cancellations are independent and similar to arrivals.

b) 22.31% probability that no cancellations will occur on a particular Wednesday

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

Mean rate of 1.5 per day on a typical Wednesday.

This means that $\mu = 1.5$

(a) Justify the use of the Poisson model.

Each wednesday is independent of each other, and each wednesday has the same mean number of cancellations.

So the answer is:

Cancellations are independent and similar to arrivals.

(b) What is the probability that no cancellations will occur on a particular Wednesday

This is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.5}*(1.5)^{0}}{(0)!} = 0.2231$

22.31% probability that no cancellations will occur on a particular Wednesday

5 0

3 years ago