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Rama09 [41]
3 years ago
6

12 more than janelle's score is 74

Mathematics
2 answers:
ankoles [38]3 years ago
5 0

Answer:

janelles score is 62

Step-by-step explanation:

kiruha [24]3 years ago
4 0

Answer:

NO

<h3><em>T</em><em>H</em><em>E</em><em> </em><em>A</em><em>N</em><em>S</em><em>W</em><em>E</em><em>R</em><em> </em><em>I</em><em>S</em><em> </em><em>7</em><em>4</em><em> </em><em>I</em><em>S</em><em> </em><em>M</em><em>O</em><em>R</em><em>E</em><em> </em><em>T</em><em>H</em><em>A</em><em>N</em><em> </em><em>1</em><em>2</em></h3>

Step-by-step explanation:

<h3><em>#</em><em>H</em><em>O</em><em>P</em><em>E</em><em> </em><em>I</em><em>T</em><em> </em><em>H</em><em>E</em><em>L</em><em>P</em><em>S</em></h3><h3><em>#</em><em>L</em><em>E</em><em>A</em><em>R</em><em>N</em><em> </em><em>M</em><em>O</em><em>R</em><em>E</em><em>,</em><em>H</em><em>E</em><em>L</em><em>P</em><em> </em><em>M</em><em>O</em><em>R</em><em>E</em></h3>
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A coin is biased such that a is times as likely to occur as a tail. Find the expected number of when this coin is tossed twice.
Sever21 [200]

Complete question:

A coin is biased such that a head is three times as likely to occur as a tail. Find the expected number of tails when this coin is tossed twice.

Answer:

1/2

Step-by-step explanation:

Head is 3 times as likely to occur as tail;

Hence,

Sample space for 2 tosses :

{HH, HH, HH, HH, HH, HH, HH, HH, HH, HT, HT, HT, HT, HT, HT, TT}

Expected probability (E(x)) = Σx*p(x)

P(x) = required outcome / Total possible outcomes

x : __________ 0 ______ 1 _______ 2

p(x) : ________9/16_____6/16_____1/16

Σx*p(x) : 0(9/16) + 1(6/16) + 2(1/16)

= 0 + 6/16 + 2/16

= 0 + 3/8 + 1/8

= 4/8

= 1/2

8 0
3 years ago
You earned $62,715 this year and your Income tax calculator says you owe $8,818 in taxes. But you neglected to deduct $1,000 for
zepelin [54]

Answer: 2,749.50

Step-by-step explanation:

56,715*10.70% = 6068.505

8818 - 6068.505 = 2,749.50

6 0
3 years ago
Which equation has a slope of –3/2 and a y-intercept of 5?
Natasha_Volkova [10]
I’m not sure what you’re options are but an equation in Y=mx+b form for this would be y= -3/2x + 5

M is always the slope
B is always the y intercept
7 0
3 years ago
Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x 2) f(x) = (x – 2)(x 2) f(x) = (x 1)(x – 2)
makvit [3.9K]

The function that has a vertex on the y-axis is f(x) = (x - 2)(x + 2)

<h3>How to determine the function?</h3>

For a function to have its vertex on the y-axis, then the coordinate of the vertex must be:

(h,k) = (0,y)

A quadratic function is represented as:

f(x) = (x - h)^2 + k

So, we have:

f(x) = (x - 0)^2 + k

Evaluate

f(x) = x^2 + k

From the list of options, we have:

f(x) = (x - 2)(x + 2)

Expand

f(x) = x^2 - 4

Hence, the function that has a vertex on the y-axis is f(x) = (x - 2)(x + 2)

Read more about vertex at:

brainly.com/question/18797214

#SPJ4

8 0
2 years ago
A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of
Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So,  y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

5 0
2 years ago
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