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IgorLugansk [536]

3 years ago

10

Which inequality has a closed circle when it is graphed on a number line? x greater-than 0.4 x greater-than negative 3 x less-th

an 4 x greater-than-or-equal-to 3

2 answers:

goldfiish [28.3K]3 years ago

5 0

Answer:

C

Step-by-step explanation:

Mice21 [21]3 years ago

4 0

Answer:

x ≥ 3

Step-by-step explanation:

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-24 points lost on a test how many were incorrect

Luba_88 [7]

if there were 100 questions then 76 were correct, and 24 incorrect, you can simplify and rearrange the fraction depending on how many questions there were,

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4 years ago

What is the standard deviation of the following data set rounded to the nearest tenth? 56 ,78 ,123 , 34, 67, 91, 20

quester [9]

Answer:

The standard deviation of the following data set is 32.2

Step-by-step explanation:

step 1

Find the mean

we have

$[56,78,123,34,67,9,20]$

Sum the data and divided by the number of elements

$[56+78+123+34+67+91+20]/7=469/7=67$

step 2

For each number: subtract the Mean and square the result

$[(56-67)^{2},(78-67)^{2},(123-67)^{2},(34-67)^{2},(67-67)^{2},(91-67)^{2},(20-67)^{2}]$

$[121,121,3,136,1,089,0,576,2,209]$

step 3

Work out the mean of those squared differences

$[121+121+3,136+1,089+0+576+2,209]/7=1,036$

This value is called the "Variance"

step 4

Take the square root of the variance

$standard\ deviation=\sqrt{1,036}=32.2$

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3 years ago

What percentage of the students

mars1129 [50]

Step-by-step explanation:

Technically if it is 10 out of 10 girls that likes pizza then that is 100%.

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3 years ago

Read 2 more answers

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

(a) The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

<u />

(b) The rate the area formed by the ladder is changing is approximately <u>-75.29 ft.²/sec</u>

<u />

(c) The rate at which the angle formed with the wall is changing is approximately <u>0.286 rad/sec</u>.

Reasons:

The given parameter are;

Length of the ladder, <em>l</em> = 25 feet

Rate at which the base of the ladder is pulled, $\displaystyle \frac{dx}{dt}$ = 2 feet per second

(a) Let <em>y</em> represent the height of the ladder on the wall, by chain rule of differentiation, we have;

$\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}$

25² = x² + y²

y = √(25² - x²)

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}$

Which gives;

$\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2$

$\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}$

When x = 15, we get;

$\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}$

The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

When x = 20, we get;

$\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6$

The velocity of the top of the ladder = $\underline{-2.\overline{6} \ m/s \ downwards}$

When x = 24, we get;

$\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86$

The velocity of the top of the ladder ≈ <u>-6.86 m/s downwards</u>

(b) $\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}$

Therefore;

$\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}$

$\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}$

$\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}$

Therefore;

$\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2$

When the ladder is 24 feet from the wall, we have;

x = 24

$\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}$

The rate the area formed by the ladder is changing, $\displaystyle \frac{dA}{dt}$ ≈ <u>-75.29 ft.²/sec</u>

(c) From trigonometric ratios, we have;

$\displaystyle sin(\theta) = \frac{x}{25}$

$\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}$

$\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}$

$\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}$

Which gives;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}$

When x = 24 feet, we have;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}$

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is $\displaystyle \frac{d \theta}{dt}$ ≈ <u>0.286 rad/sec</u>

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

3 0

3 years ago

With a rectangle length of 309 cm.and width of 249cmdoes it have the perimeter of 1500 cm?

horsena [70]

1116 is the perimeter it has. The formula for perimeter is 2l+2w=p
so when you plug the numbers in you get 2(309)+2(249)=1116. So it does not have a perimeter of 1500 cm.

7 0

4 years ago