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3 years ago

9

The temperature is -14 degrees. Over the next three days, the temperature drops 5 degrees, then 3 degrees, and finally 1 degree.

What is the temperature now?

2 answers:

Margaret [11]3 years ago

5 0

Answer:

-23 degree

Step-by-step explanation:

=>-14-5-3-1

=>-23

ivann1987 [24]3 years ago

3 0

Answer:

-23°

Step-by-step explanation:

= -14 - (5+3+1)

= -14 - 9

= -23°

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Answer:

The probability that on a given day below 6,214,323 drinks are sold is 0.8413.

Step-by-step explanation:

The provided information are:

Population mean $(\mu)$ = 6,205,195

Population standard deviation $(\sigma)$ = 9,120.32

Consider, <em>X</em> be the random variable that represents the number of drinks that are sold on a given that is normally distributed with the mean = 6,205,195 and standard deviation = 9,120.32.

The probability that on a given day below 6,214,323 drinks are sold can be calculated as:

$P(X$

It must be noted that <em>P</em>(<em>Z</em> < 1.0008) = 0.8413 has been calculated using the standard normal table.

Hence, the required probability is 0.8413.

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3 years ago

Mr. Lott has 250 more centimeters of tape than Mrs. Sanchez. Mr. Lott has 775 centimeters of tape. How many centimeters of tape

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Step-by-step explanation:

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3 years ago

If P=(3,4), find:<br> R y-axis (P)<br> Please help!!

user100 [1]

Answer:

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Step-by-step explanation:

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4 years ago

Read 2 more answers

You have 108 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the River

Licemer1 [7]

Answer:

The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.

Step-by-step explanation:

Given that,

The length of fencing of the rectangular plot is = 108 ft.

Let the longer side of the rectangular plot be x which is also the side along the river side and the width of the rectangular plot be y.

Since the fence along the river does not need.

So the total perimeter of the rectangle is =2(x+y) -x

=2x+2y-y

=x+2y

So,

x+2y =108

⇒x=108 -2y

Then the area of the rectangle plot is A = xy

A=xy

⇒A= (108-2y)y

⇒ A = 108y-2y²

A = 108y-2y²

Differentiating with respect to x

A'= 108 -4y

Again differentiating with respect to x

A''= -4

For maximum or minimum, A'=0

108 -4y=0

⇒4y=108

$\Rightarrow y=\frac{108}{4}$

⇒y=27.

$A''|_{y=27}=-4$

Since at y= 27, A''<0

So, at y=27 ft , the area of the rectangular plot maximum.

Then x= (108-2.27)

=54 ft.

The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.

3 0

3 years ago