All categories

2 years ago

Plz help due soon real answers plz

Mathematics

2 answers:

Keith_Richards [23]2 years ago

6 0

The answer is 3 feet. have a nice day

maksim [4K]2 years ago

3 0

Your answer could be 3 feet

You might be interested in

Use the drawing tools to form the correct answers on the graph. Plot the vertex and the axis of symmetry of this function: f(x)

CaHeK987 [17]

Answer:

Axis of Symmetry: x = 3

Vertex: (3, 5)

Step-by-step explanation:

Use a graphing calc.

6 0

3 years ago

Read 2 more answers

The temperature changes can be represented by the expression -0.5( 1 -2h).Chase says that the temperature change can also be rep

UNO [17]

Answer:

The expression are equivalent

Step-by-step explanation:

The question requests that we show whether the equation -0.5( 1 -2h) is equivalent to 4(2h + 2) - 7h - 8.5 or not :

Simplifying 4(2h + 2) - 7h - 8.5 ; we have ;

4(2h + 2) - 7h - 8.5

8h + 8 - 7h - 8.5

8h - 7h + 8 - 8.5

h - 0.5

-0.5(1 - 2h)

-0.5 + h

= h - 0.5

Hence, both expressions are the same

3 0

3 years ago

error uses 1/3 can of wet dog food for his dog,muddy,each day.how many servings will he get from 5 cans of dog food?

Vesnalui [34]

First convert 5 cans of dog food to a fraction
(5*3)/3 = 15/3
15/3 - 1/3 = 14/3
he uses 1/3 a can and so he then has 14 servings.

7 0

2 years ago

Messages arrive to a computer server according to a Poisson distribution with a mean rate of 11 per hour. Determine the length o

Zarrin [17]

Answer:

The length of the interval during which no messages arrive is 90 seconds long.

Step-by-step explanation:

Let X = number of messages arriving on a computer server in an hour.

The mean rate of the arrival of messages is, λ = 11/ hour.

The random variable X follows a Poisson distribution with parameter λ = 11.

The probability mass function of X is:

$P(X=x)=\frac{e^{-11}11^{x}}{x!};\ x=0,1,2,3....$

It is provided that in t hours the probability of receiving 0 messages is,

P (X = 0) = 0.76

Compute the value of t as follows:

$P(X=0)=\frac{e^{-11\times t}(11\times t)^{0}}{0!}\\0.76=e^{-11\times t}\\\ln(0.76)=-11\times t\\t=-\frac{\ln(0.76)}{11} \\=0.025\ hours\\\approx90\ seconds$

Thus, the length of the interval during which no messages arrive is 90 seconds long.

6 0

3 years ago

Please help me! I really need it. ...You deposit 1500 every month starting today in an investment account but would withdraw 750

photoshop1234 [79]

Answer: multiply 750 and 12

Divide that by the 14%

Step-by-step explanation:

5 0

2 years ago