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Dimas [21]
3 years ago
8

A number, x, rounded to 2 significant figures is 1300Write down the error interval for x.​

Mathematics
1 answer:
nikitadnepr [17]3 years ago
7 0

Answer:

[1250,1350).

Step-by-step explanation:

It is given that a number, x, rounded to 2 significant figures is 1300.

It is possible if,

1. The value of x is greater than of equal to 1250 and less than or equal to 1300.

i.e., 1250\leq x\leq 1300     ...(1)

2. The value of x is greater than of equal to 1300 and less than 1350.

i.e., 1300\leq x       ...(2)

On combining (1) and (2), we get

1250\leq x < 1350

1350 is not included in the error interval for x.​

Interval notation is [1250,1350).

Therefore, the error interval for x is [1250,1350).

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3 years ago
Koran is spending $12 on games and rides at another carnival, where a game costs $0.25 and a ride costs $1.
Vlad [161]

Answer:

0.25x+y=12

Step-by-step explanation:

Let

y ----> the number of rides

x ---> the number of games

we know that

The number of rides multiplied by  $1 plus the number of games multiplied by $0.25 must be equal to $12 (the dollar amount Koran is spending)

so

The equation that represent this situation is

0.25x+y=12

5 0
3 years ago
Liam is going to a carnival that has games and rides. Each game costs $5 and each ride costs $3.50. How much would Liam have to
lesya692 [45]

Answer:

$43

Step-by-step explanation:

You have to find how much he is paying for his games and how much he is paying for his rides, then add them together.

5(3) = $15 (games)

8(3.5) = $28 (rides)

28+15 = 43

5 0
3 years ago
Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
fredd [130]

f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

5 0
3 years ago
-x + y = -5 Y intersect and x intersect
Furkat [3]

Answer:

  • (0, - 5) and (5, 0)

Step-by-step explanation:

<u>Given line:</u>

  • - x + y = - 5

<u />

<u>The y-intercept is the point when x = 0:</u>

  • 0 + y = - 5
  • y = - 5

The point is (0, - 5)

<u />

<u>The x-intercept is the point when y = 0:</u>

  • - x + 0 = - 5
  • - x = - 5
  • x = 5

The point is (5, 0)

5 0
3 years ago
Read 2 more answers
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