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nikdorinn [45]
3 years ago
14

Helllps please help this timed please I will mark brainliest. PLz. NO LINKS OR I WILL REPORT YOU. Please give an explanation.

Mathematics
1 answer:
USPshnik [31]3 years ago
3 0

Answer:

please????

Step-by-step explanation:

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Consider the graph of the function f(x)=2x.
77julia77 [94]

Given:

The parent function is:

f(x)=2^x

The other function is:

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To find:

The statement that describes a key feature of function g.

Solution:

We have,

f(x)=2^x

g(x)=2f(x)

Using these two functions, we get

g(x)=2(2)^x

Putting x=0, we get

g(x)=2(2)^{(0)}

g(x)=2(1)

g(x)=2

The y-intercept of the function g at (0,2). So, option A is correct and option B is incorrect.

We know that g(x)\to 0 as x\to -\infty and it will never intersect the line y=0. It means the horizontal asymptote of the function g is

Therefore, the correct option is A.

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3 years ago
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What is the probability that a randomly selected person is 18 to 34 years old GIVEN that they are in favor of increasing the min
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Step-by-step explanation:

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3 years ago
A tennis ball is dropped from a height of 12 meters. Each time the ball bounces back to 80% of the height from which it fell. Dr
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3 years ago
Multiply radicals please​
leva [86]

Answer:

B

Step-by-step explanation:

~~~\left( \dfrac 1 3 \sqrt x - \dfrac 32 \sqrt y \right)^2\\\\\\=\left( \dfrac 13 \sqrt x \right)^2 + \left( \dfrac 32 \sqrt y \right)^2 - 2 \left(\dfrac 13 \sqrt x \right)\left( \dfrac 32 \sqrt y \right)\\\\\\=\dfrac 19 x+ \dfrac 94 y-\sqrt{xy}\\\\\\=\dfrac 19x - \sqrt{xy} + \dfrac 94 y

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2 years ago
1) Find the 95 % confidence interval for the mean if σ = 11 , using this sample:
melamori03 [73]

Answer:

The 95 % confidence interval for the mean is between 28.09 and 35.97.

Step-by-step explanation:

Mean of the sample:

30 values.

So we sum all the values and divide by 30.

The sum of all the values(43 + 52 + 18 + ... + 20 + 41) is 961.

961/30 = 32.03

Confidence interval:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96\frac{11}{30} = 3.94

The lower end of the interval is the sample mean subtracted by M. So it is 32.03 - 3.94 = 28.09

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The 95 % confidence interval for the mean is between 28.09 and 35.97.

8 0
3 years ago
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