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DanielleElmas [232]

2 years ago

11

Students who party before an exam are twice as likely to fail as those who don't party (and presumably study). If 20% of the stu

dents partied before the exam, what fraction of the students who failed went partying? unanswered

1 answer:

True [87]2 years ago

8 0

Answer:

The fraction of the students who failed to went partying = $\frac{1}{10}$

Step-by-step explanation:

Let total number of students = 100

No. of students partied are twice the no. of students who not partied.

⇒ No. of students partied = 2 × the no. of students who are not partied

No. of students partied before the exam = 20 % of total students

⇒ No. of students partied before the exam = $\frac{20}{100}$ × 100

⇒ No. of students partied before the exam = 20

No. of students who not partied before the exam = $\frac{20}{2} = 10$

Thus the fraction of the students who failed to went partying = $\frac{10}{100} = \frac{1}{10}$

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Check all statements that are equivalent to If the sky is not clear, then you don't see the stars ? a. Clear sky is necessary an

IrinaVladis [17]

Answer: The answer is G

Step-by-step explanation:

If you don't see the stars, then the sky is not clear. This statement is equivalent to saying "If the sky is not clear, then you don't see the stars", because if you rearrange both statements you will find out that they are equivalent to each other.

Mark out all the words in each statement, you will discover that at the end you will have no word left on both side, its just as good as saying 1 - 1 = 0.

6 0

2 years ago

Read 2 more answers

4. A cruise ship travels in the direction of 55degrees for 40 miles, then changes course to a direction of 100 degrees for 35 mi

olga2289 [7]

Answer:

69.3 mi

Step-by-step explanation:

Let x represent the distance of the ship from its original position.

x²= 40² + 35² -2(40)(35)cos(135)

$x^{2} =4804.9$

$\sqrt{x} ^{2} = \sqrt{4804.9}$

x= 69.3 mi

8 0

1 year ago

Raymond just got done jumping at Super Bounce Trampoline Center. The total cost of his session was \$43.25$43.25dollar sign, 43,

Elodia [21]

Answer:

29 minutes

Step-by-step explanation:

Given

$Total = \$43.25$

$Entrance\ Fee = \$7$

$Per\ Minute = \$1.25$

Required

Determine the number of minutes (t) that he was on the trampoline.

This can be expressed as:

$Total\ Payment = Entrance\ Fee + Amount\ per\ minute * t$

Where t represents the number of minutes

Substitute values for Total, Entrance Fee and Amount per minute

$43.25 = 7 + 1.25 * t$

$43.25 = 7 + 1.25t$

Solving further to get the value of t, we have:

$1.25t = 43.25 - 7$

$1.25t = 36.25$

$t = 36.25/1.25$

$t = 29$

3 0

1 year ago

Read 2 more answers

Triangles E F G and E prime F prime G prime are shown. Angles F E G and F prime E prime G prime are 72 degrees. Angles E F G and

Karolina [17]

Answer:

We need a non-included side of one triangle

Step-by-step explanation:

By means of the AAS postulate.

The Angle-Angle-Side postulate (AAS) tells us that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the two triangles are congruent.

3 0

1 year ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1x2+y2≤1. the plate, including t

vredina [299]

You're looking for the extreme values of $x^2+3y^2+13x$ subject to the constraint $x^2+y^2\le1$ .

The target function has partial derivatives (set equal to 0)

$\dfrac{\partial(x^2+3y^2+13x)}{\partial x}=2x+13=0\implies x=-\dfrac{13}2$

$\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0$

so there is only one critical point at $\left(-\dfrac{13}2,0\right)$ . But this point does not fall in the region $x^2+y^2\le1$ . There are no extreme values in the region of interest, so we check the boundary.

Parameterize the boundary of $x^2+y^2\le1$ by

$x=\cos u$

$y=\sin u$

with $0\le u$ . Then $t(x,y)$ can be considered a function of $u$ alone:

$t(x,y)=t(\cos u,\sin u)=T(u)$

$T(u)=\cos^2u+3\sin^2u+13\cos u$

$T(u)=3+13\cos u-2\cos^2u$

$T(u)$ has critical points where $T'(u)=0$ :

$T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0$

$(1)\quad\sin u=0\implies u=0,u=\pi$

$(2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4$

but $|\cos u|\le1$ for all $u$ , so this case yields nothing important.

At these critical points, we have temperatures of

$T(0)=14$

$T(\pi)=-12$

so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.

4 0

1 year ago