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2 years ago

In a group of 39 students, 14 study both Art and Biology.

Mathematics

1 answer:

Blababa [14]2 years ago

3 0

Answer:

66.666666666667%

Step-by-step explanation:

26 study biology

26/39= .66666666666667

convert into a percent

100 times .66666666666667 =66.666666666667%

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Pete’s grade on a test was within 5 points of his class average of 94. What is his range of grades on the test?

d1i1m1o1n [39]

I'm pretty sure it would be 89-99.

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3 years ago

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a rectangle garden is 2 inches by 1.5 inches in a drawing the scale is 1 inch= 30 yards what is the actual area of the garden

dolphi86 [110]

If 1 inch = 30 yards, 2 inches is 60 yards and 1.5 inches is 45 yards. This means the area of the garden is 60 * 45 or 2700 yards

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2 years ago

Question 2 (5 points)

IrinaK [193]

Answer:

$x - y + 5 = 0$

Step-by-step explanation:

Given:

Line passes through (-2,3) and (2,7)

Required:

Find the equation of the line

The equation of the line passing through two points is given by

$\frac{y-y_{1} }{y_{2}- y_{1} } =\frac{x_{}- x_{1} }{x_{2} -x_{1} }$

So, $\frac{y-3}{7-3} =\frac{x-(-2))}{2-(-2)}$

$\frac{y-3}{4} =\frac{x+2}{2+2}$

$\frac{y-3}{4} =\frac{x+2}{4}$

Now, $y-3 = x+2$

On rearranging,

$x - y + 3 + 2 = 0x - y + 5 = 0$

Therefore, the equation of the line is $x - y + 5 = 0.$

5 0

2 years ago

What is the mean absolute deviation? 3 7 6 4 7 9

Karo-lina-s [1.5K]

Answer:

1.6666666666667

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2 years ago

The cable between two towers of a suspension bridge can be modeled by the function shown, where x and y are measured in feet. Th

hodyreva [135]

a) x = 200 ft

b) y = 50 ft

Domain: $0\leq x \leq 400$

Range: $50\leq y \leq 150$

Step-by-step explanation:

The function that models the carble between the two towers is:

$y=\frac{1}{400}x^2-x+150$

where x and y are measured in feet.

Here we want to find the lowest point of the cable: this is equivalent to find the minimum of the function $y(x)$ .

In order to find the minimum of the function, we have to calculate its first derivative and require it to be zero, so:

$y'(x)=0$

The derivative of y(x) is:

$y'(x)=\frac{1}{400}\cdot 2 x^{2-1}-1=\frac{1}{200}x-1$

And requiring it to be zero,

$\frac{1}{200}x-1=0$

Solving for x,

$\frac{1}{200}x=1\\x=200 ft$

In order to find how high is the road above the water, we have to find the value of y (the height of the cable) at its minimum value, because that is the point where the cable has the same height above the water as the road.

From part a), we found that the lowest position of the cable is at

$x=200 ft$

If we now substitute this value into the expression that gives the height of the cable,

$y=\frac{1}{400}x^2-x+150$

We can find the lowest height of the cable above the water:

$y=\frac{1}{400}(200)^2-200+150=50 ft$

Therefore, the height of the road above the water is 50 feet.

Domain:

The domain of a function is the set of all possible values that the independent variable x can take.

In this problem, the extreme points of the domain of this function are represented by the position of the two towers.

From part a), we calculated that the lowest point of the cable is at x = 200 ft, and this point is equidistant from both towers. If we set the position of the tower on the left at

$x=0 ft$