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Flauer [41]
3 years ago
14

Solve for x. 2.5x=50

Mathematics
2 answers:
kompoz [17]3 years ago
7 0

Answer:

x=20

Step-by-step explanation:

2.5x=50

\frac{2.5}{2.5} =\frac{50}{2.5}

x = 20

Have a great day!!!!

PumpkinSpice1

Doss [256]3 years ago
4 0

Answer:

bopbeep

Step-by-step explanation:

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pishuonlain [190]

Answer:

D is the answers for the question

Step-by-step explanation:

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3 years ago
Please I need help I don't understand
mario62 [17]

Answer:

Step-by-step explanation:

from the graph of f(x)

when x=1,f(x)=0

or f(1)=0

when f(x)=2,x=2

for g(x)

when x=6,g(x)=16

or g(6)=16

when g(x)=18,x=32

for h(x)

when x=14

h(x)=27x-7

h(14)=27×14-7=7(27×2-1)=7(54-1)=7×53=371

h(x)=-493

27x-7=-493

27 x=-493+7=-486

3 x=-54

x=-18

for p(t)

when t=94

p(t)=24

p(94)=24

p(t)=67

t=31

8 0
3 years ago
Y is a random variable that is distributed N(-16, 1.21). Find k such that Prob(-15.043 &lt; Y ≤ k) = 0.1546. (Round your answer
IgorLugansk [536]

Transform <em>Y</em> to <em>Z</em>, which is distributed N(0, 1), using the formula

<em>Y</em> = <em>µ</em> + <em>σZ</em>

where <em>µ</em> = -16 and <em>σ</em> = 1.21.

Pr[-15.043 < <em>Y</em> ≤ <em>k</em>] = 0.1546

Pr[(-15.043 + 16)/1.21 < (<em>Y</em> + 16)/1.21 ≤ (<em>k</em> + 16)/1.21] = 0.1546

Pr[0.791 < <em>Z</em> ≤ (<em>k</em> + 16)/1.21] ≈ 0.1546

Pr[<em>Z</em> ≤ (<em>k</em> + 16)/1.21] - Pr[<em>Z</em> < 0.791] = 0.1546

Pr[<em>Z</em> ≤ (<em>k</em> + 16)/1.21] = 0.1546 + Pr[<em>Z</em> < 0.791]

Pr[<em>Z</em> ≤ (<em>k</em> + 16)/1.21] ≈ 0.1546 + 0.786

Pr[<em>Z</em> ≤ (<em>k</em> + 16)/1.21] ≈ 0.940

Take the inverse CDF of both sides (<em>Φ(x)</em> denotes the CDF itself):

(<em>k</em> + 16)/1.21 ≈ <em>Φ⁻¹</em> (0.940) ≈ 1.556

Solve for <em>k</em> :

<em>k</em> + 16 = 1.21 • 1.556

<em>k</em> ≈ -14.118

8 0
3 years ago
Please help 5x^2+21x+4
SVEN [57.7K]

Answer:

(5x+1) (x+4)

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Can someone answer fast.plsss
Anika [276]

Answer:

a

Step-by-step explanation:

sec (thita) = squrt (5)

squaring on both sides:

sec^2 (thita) = 5 - equation 1

1 + tan^2 (thita) = 5

tan^2 (thita) = 4

tan (thita) = 2.

= tan (thita) - squrt(5)sin(thita)

= 2 - squrt(5) x 2/ squrt(5)

= 0

from eqn - 1

sec(thita) = squrt(5)

cos(thita) = 1/ squrt(5)

sin(thita) = squrt ( 1- 1/(squrt (5))^2)

sin(thita) = 2/ squrt(5) .

4 0
3 years ago
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