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Svetllana [295]
3 years ago
8

Solve for x

}^{2} = 7x + 2" alt="4 {x}^{2} = 7x + 2" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Alina [70]3 years ago
7 0

\bf 4x^2=7x+2\implies 4x^2-7x-2=0\implies (4x+1)(x-2)=0 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 4x\cdot x=\boxed{4x^2}~\hfill (1)(-2)=\boxed{-2}\hfill (4x\cdot -2)+(1\cdot x)=\boxed{-7x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} 4x+1=0\implies 4x=-1\implies &x=-\cfrac{1}{4}\\[1em] x-2=0\implies &x=2 \end{cases}

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d+p=850\\300d+225p=217500\\\\300d+300p=255000\\300d+225p=217500\\75p=37500\\p=500\\d+(500)=850\\d=350

350 Douglas firs and 500 Ponderosa pines.
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3 years ago
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4.20. According to a report released by the National Center for Health Statistics, 51% of U.S. households has only cell phones (
Sergeu [11.5K]

Answer:

The probability that the household has only cell phones and has high-speed Internet is 0.408

Step-by-step explanation:

Let A be the event that represents U.S. households has only cell phones

Let B be the event that represents U.S. households have high-speed Internet.

We are given that 51% of U.S. households has only cell phones

P(A)=0.51

We are given that 70% of the U.S. households have high-speed Internet.

P(B)=0.7

We are given that U.S. households having only cell phones, 80% have high-speed Internet. A U.S household is randomly selected.

P(B|A)=0.8

\frac{P(A\capB)}{P(A)}=0.8\\P(A\capB)=0.8 \times P(A)\\P(A\capB)=0.8 \times 0.51\\P(A\capB)=0.408

Hence the probability that the household has only cell phones and has high-speed Internet is 0.408

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3 years ago
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Answer:

y-values of sampled wave:

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Step-by-step explanation:

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3 years ago
I need help ASAP! It's urgent.. PLISSSSS​
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Answer:

a) 6 mins

b) 70km/h

c) t= 45

Step-by-step explanation:

a) The bus stops from t=10 to t=16 minutes since the distance the busvtravelled remained constant at 15km

Duration

= 16 -10

= 6 minutes

b) Average speed

= total distance ÷ total time

Total time

= 24min

= (24÷60) hr

= 0.4 h

Average speed

= 28 ÷0.4

= 70 km/h

c) Average speed= total distance/ total time

Average speed

= 80km/h

= (80÷60) km/min

= 1⅓ km/min

1⅓= 28 ÷(t -24)

<em>since</em><em> </em><em>duration</em><em> </em><em>for</em><em> </em><em>return</em><em> </em><em>journey</em><em> </em><em>is</em><em> </em><em>from</em><em> </em><em>t</em><em>=</em><em>2</em><em>4</em><em> </em><em>mins</em><em> </em><em>to</em><em> </em><em>t</em><em> </em><em>mins</em><em>.</em>

\frac{4}{3}(t -24)= 28

\frac{4}{3}t - 32= 28

\frac{4}{3}t= 32 +28

\frac{4}{3}t= 60

t= 6 0\div  \frac{4}{3}

t= 45

*Here, I assume that this is a displacement- time graph, so the distance shown is the distance of the bus from the starting point because technically if it is a distance-time graph, the distance would still increase as the bus travels the 'return journey'.

Thus, distance is decreasing after t=24 and reaches zero at time= t mins so that is the return journey. (because when the bus returns back to starting point, displacement/ distance from starting point= 0km)

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Answer: I can answer 20

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