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LenKa [72]
2 years ago
15

There are 80 students in a school. Every student is taking either algebra or pre-calculus, and some students are taking both. Th

ere are 60 students in the algebra class. If 15 students are taking both algebra and pre-calculus, what is the probability that a randomly selected student is taking pre-calculus but not algebra?
Mathematics
1 answer:
Finger [1]2 years ago
3 0

Answer:

1/4

Step-by-step explanation:

Total students = 80

Students taking both classes =  15

Students taking algebra class = 60

Students taking only algebra class = 60 - 15 = 45

Students taking only pre-calculus class = 80- (45+15) = 80 - 60 = 20

P( students taking pre-calculus but not algebra) = 20/80 = 1/4

                         

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Jillian is trying to save water, so she reduces the size of her square grass lawn by 8 feet on each side. The area of the smalle
siniylev [52]

Answer:

20 by 20

Step-by-step explanation:

the new dimensions have to be 12 because 12×12 =144 so you would add 8 to 12 and get 20 for each side

7 0
2 years ago
Read 2 more answers
What are the zeros of the quadratic function f(x) = 6x^2 + 12x – 7?
strojnjashka [21]

Answer:

x1=\frac{-2+\sqrt{26/3}}{2}

x2=\frac{-2-\sqrt{26/3} }{2}

Step-by-step explanation:

To find the zeros of the quadratic function f(x)=6x^2 + 12x – 7 we need to factorize the polynomial.

To do so, we need to use the quadratic formula, which states that the solution to any equation of the form ax^2 + bx + c = 0 is:

x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

So, the first thing we're going to do is divide the whole function by 6:

6x^2 + 12x – 7 = 0 -> x^2 + 2x - 7/6

This step is optional, but it makes things quite easier.

Then we using the quadratic formula, where:

a=1, b= 2, c = -7/6.

Then:

x=\frac{-2±\sqrt{2^{2}-4(1)(-7/6)}}{2}

x=\frac{-2±\sqrt{4 +14/3}}{2}

x=\frac{-2±\sqrt{26/3}}{2}

So the zeros are:

x1=\frac{-2+\sqrt{26/3}}{2}

x2=\frac{-2-\sqrt{26/3}}{2}

4 0
2 years ago
Geoffrey is evaluating the expression StartFraction (negative 3) cubed (2 Superscript 6 Baseline) Over (Negative 3) Superscript
makkiz [27]

The mathematical expression does not seem clear but I have made an attempt to make sense of what is implied.

Answer:

<em>a</em> = 4, <em>b</em> = 2, <em>c</em> = 16, <em>d</em> = 9

Step-by-step explanation:

\dfrac{(-3)^3(2^6)}{(-3)^5(2^2)} = \dfrac{(2)^a}{(-3)^b} = \dfrac{c}{d}

Solving the first part of the question by indices,

\dfrac{(-3)^3(2^6)}{(-3)^5(2^2)} = (-3)^{3-5}(2)^{6-2} = (-3)^{-2}(2)^{4} = \dfrac{(2)^4}{(-3)^2}

Comparing the rightmost term with the second term in the question,

<em>a</em> = 4, <em>b</em> = 2

Solving on,

\dfrac{(2)^4}{(-3)^2} = \dfrac{(2)\times(2)\times(2)\times(2)}{(-3)\times(-3)} = \dfrac{16}{9}

Comparing with the final term in the question,

<em>c</em> = 16 and <em>d</em> = 9

Therefore,

<em>a</em> = 4, <em>b</em> = 2, <em>c</em> = 16, <em>d</em> = 9

3 0
2 years ago
Read 2 more answers
The speed of light is about 1.86×10^5 miles per second. The star Sirius is about 5.062×10^13 miles from Earth. About how many se
marysya [2.9K]

Answer:

2.72 × 10^8 seconds

Step-by-step explanation:

1. Organize the given information to help you create the equation:

- Light travels 1.86 × 10^5 miles per 1 second

- Light travels 5.062 × 10^13 miles per <em>n</em> seconds

- Need to solve for <em>n</em>

2. Set up the equation (Make sure that both numerators and both denominators match units, so the seconds are either both on top or both on bottom, NOT switched):

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3. Cross multiply (Multiply the first fraction's numerator by the second's denominator, and vice versa. It doesn't matter what number is on which side of the = sign)

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4. Solve for <em>n</em> by isolating the variable and dividing.

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<em>n</em> = \frac{5.062 * 10^1^3}{1.86 * 10^5}

<em>n </em>= 2.72 × 10^8 (Make sure to include the unit, seconds)

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2 years ago
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tankabanditka [31]
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2 years ago
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