Can you please help me with this question?

Pls help with the explanation and answer ty!

Zigmanuir [339]

I’m sorry, the text is too small. But I hope you find your answer soon.

6 0

3 years ago

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Find the tangent line equation of the curve at the given point. Y=arcsin(7x) at the point where x=sqrt2/14

Mumz [18]

Answer:

$Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

Step-by-step explanation:

The equation of the curve is

$Y = sin^{-1}(7x)$

To find the equation of tangent we need to differentiate this equation w.r.t x

So, differentiating we get

$Y'=\frac{7}{\sqrt{1-49x^2} }$

This would give the slope of the tangent line at any given point of which x coordinate is known. In the present case it is $x = \sqrt{\frac{1}{7} }$

Then slope would accordingly be

$Y'=\frac{7}{\sqrt{1-49/49} }$

= ∞

For, $x = \sqrt{\frac{1}{7} }$ , $Y = sin^{-1}(7/7)= \pi/2$

Equation of tangent line, in the point slope form, would be $Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

4 0

3 years ago

For an activity in class, student used pieces of string to measure road on a map. They used a map scale to determine the distanc

ivanzaharov [21]

Answer:

Step-by-step explanation:

85miles¯\_(ツ)_/¯

7 0

3 years ago

In 1996 ato boldon of ucla ran the 100-meter dash in 9.92 seconds. in 1969 john carlos of san jose state ran the 100-yard dash i

Neko [114]

There is a app called Cymath it shows you how to do it and gives you the answer

3 0

3 years ago

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A random sample of n measurements was selected from a population with unknown mean mu and standard deviation sigmaequals50 for e

Andre45 [30]

Answer:

a) (26.50;57.50)

b) (117.34;128.66)

c) (12.13;27.87)

d) (-4.73;11.01)

e) No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval is given by this formula:

$\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ (1)

And for a 95% of confidence the significance is given by $\alpha=1-0.95=0.05$ , and $\frac{\alpha}{2}=0.025$ . Since we know the population standard deviation we can calculate the critical value $z_{0.025}= \pm 1.96$

Part a

$n=40,\bar X=42,\sigma=50$

If we use the formula (1) and we replace the values we got:

$42 - 1.96 \frac{50}{\sqrt{40}}=26.50$

$42 + 1.96 \frac{50}{\sqrt{40}}=57.50$

The 95% confidence interval is given by (26.50;57.50)

Part b

$n=300,\bar X=123,\sigma=50$

If we use the formula (1) and we replace the values we got:

$123 - 1.96 \frac{50}{\sqrt{300}}=117.34$

$123 + 1.96 \frac{50}{\sqrt{300}}=128.66$

The 95% confidence interval is given by (117.34;128.66)

Part c

$n=155,\bar X=20,\sigma=50$

If we use the formula (1) and we replace the values we got:

$20 - 1.96 \frac{50}{\sqrt{155}}=12.13$

$20 + 1.96 \frac{50}{\sqrt{155}}=27.87$

The 95% confidence interval is given by (12.13;27.87)

Part d

$n=155,\bar X=3.14,\sigma=50$

If we use the formula (1) and we replace the values we got:

$3.14 - 1.96 \frac{50}{\sqrt{155}}=-4.73$

$3.14 + 1.96 \frac{50}{\sqrt{155}}=11.01$

The 95% confidence interval is given by (-4.73;11.01)

Part e

No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

8 0

3 years ago

Can you please help me with this question?​

Can you please help me with this question?