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2 years ago

The range of which function is (2, infinity)? y = 2x y = 2(5x) y = 5x +2 y = 5x + 2

Mathematics

2 answers:

Arturiano [62]2 years ago

5 0

Answer:

y = 5∧× + 2

Step-by-step explanation:

Sveta_85 [38]2 years ago

3 0

Answer:

I think your functions are $y=2^{x}$ , $y=2*5^{x}$ and $y=2+5^{x}$

If yes then then the third function which is $y=2+5^{x}$ .

Step-by-step explanation:

The function $c^{x}$ where c is a constant has

Domain : $c\geq 0$

Range : ( 0 , ∞ )

The above range is irrespective of the value of c.

I have attached the graph of each of the function, you can look at it for visualization.

$y=2^{x}$ ⇒ This function is same as $c^{x}$ so its range is ( 0 , ∞ ).

$y=2*5^{x}$ ⇒ If we double each value of the function $y=5^{x}$ , which has range ( 0 , ∞ ), but still the value of extremes won't change as 0*2=0 and ∞*2=∞. Therefore the range remains as ( 0 , ∞ ).

$y=2+5^{x}$ ⇒ If we add 2 to each value of the function $y=5^{x}$ , which has range ( 0 , ∞ ), the lower limit will change as 0+2=2 but the upper limit will be same as ∞. Therefore the range will become as ( 2 , ∞ ).

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8. Peter and his partner are conducting a physics experiment on pendulum motion. Their 30-cm

katrin2010 [14]

Answer: 90/pi degrees

Step-by-step explanation:

It forms a 15cm arc from a circle of radius 30 cm.

The diameter is 30*2*pi = 60pi cm. So the arc is 15/60pi = 1/4pi of the way around a 360 degree circle. This is 1/4pi * 360 = 90/pi degrees.

Hope that helped,

-sirswagger21

8 0

2 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

2 years ago

In a basket, 5/9 of the fruits are apples and the rest are oranges. 3/10 of the apples are green in colour. There are 15 green a

ra1l [238]

In a basket, 5/9 of the fruits are apples and the rest are oranges. 3/10 of the apples are green in colour. There are 15 green apples. How many fruits are there in the basket?