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scoray [572]
2 years ago
8

it costs 13$ for admission to amusement park, plus 1.50$ for each ride.if you have a total of 35.50 to spend, what is the greate

st number of rides you can go on
Mathematics
1 answer:
ICE Princess25 [194]2 years ago
3 0

Answer:

<em>Thus, the greatest number of rides I can go on is 15 rides</em>

Step-by-step explanation:

I have $35.50 to spend for admission plus rides in the amusement park.

Admission costs $13. That leaves me with

$35.50 - $13 = $22.50

to spend on rides.

Considering each ride costs $1.50, we can find the maximum number of rides by dividing the remaining money by the cost per ride as follows:

$22.50 / $1.50 = 15

Thus, the greatest number of rides I can go on is 15 rides

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The random variable X~(30,2^2) <br> Find p(X&lt;33) <br> Find p(X&gt;26)
Ostrovityanka [42]

Answer:

i) P(X<33)  = 0.9232

ii) P(X>26) = 0.001

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the mean of the Population = 30

Given that the standard deviation of the Population = 4

Let 'X' be the Normal distribution

<u>Step(ii):-</u>

i)

Given that the random variable  X = 33

               Z = \frac{x-mean}{S.D}

               Z = \frac{33-30}{2} = 1.5 >0

P(X<33) = P( Z<1.5)

              = 1- P(Z>1.5)

             = 1 - ( 0.5 - A(1.5))

             = 0.5 + 0.4232

  P(X<33)  = 0.9232

<u>Step(iii) :-</u>

Given that the random variable  X = 26

               Z = \frac{x-mean}{S.D}

               Z = \frac{33-26}{2} = 3.5 >0

P(X>26)  = P( Z>3.5)

              = 0.5 - A(3.5)

              = 0.5 - 0.4990

             = 0.001

P(X>26) = 0.001

5 0
2 years ago
143+ [(96 + 8) 5] – 42.
Tresset [83]

Answer:

621

Explanation:

143 + (96 + 8)(5) − 42

(96 + 8)(5)

= 520

143 + 520 - 42

= 663

663 - 42

= 621

7 0
3 years ago
One way to increase acceleration is by
Hunter-Best [27]

Answer:

I believe its D , to increase the acceleration of a object you have  to increase the force  preferably ,the mass kind of slows it down because of the weight .

But d is the most legit answer so  :p

hoped I helped -

Sleepy~

5 0
2 years ago
Plssssss help fast!!!!Launching From a Tower
aleksandrvk [35]

Answer:

A. h = 64

B.  t = 4 -> 4 seconds

C. t = 1.5 -> 1.5 seconds

D. maximum height is 100 ft

E. The domain that makes sense for the function in this context is t

is any positive real number since time can not be negative.

Step-by-step explanation:

h = -16t2 + vt + 64

A. What tower platform height was the projectile launched from?

when the projectile was not launched, t = 0

h = -16(0)^2 + v(0) + 64 = 64

B. How long was the projectile in the air?

if the projectile lands, its height = 0 so substitute 0 for h

0 = -16(t)^2 + 48(t) + 64

  = -16(t^2 - 3t -  4)

 = -16(t - 4)(t + 1)

t = 4 or t = -1

Since time can not be negative, t = -1 can not be the answer. Therefore, the projectile lands when t = 4 or 4 seconds.

C. When did it reach its maximum height?

maximum -> t=-b/2a where in -16(t)^2 + 48(t) + 64, b = 48 and a = -16

t = -48/-32 = 1.5

D. What was its maximum height?

plug t = 1.5 into -16(t)^2 + 48(t) + 64

-16(1.5)^2 + 48(1.5) + 64 = -36 + 72 + 64 = 100

E. The domain that makes sense for the function in this context is t

is any positive real number since time can not be negative.

6 0
2 years ago
Pleaseeeeeeees help meee
maria [59]
12, 4 is the answer.
8 0
2 years ago
Read 2 more answers
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