A plane left Atlanta at 4:50 and arrived at Boston in 3 hours and 15 minutes. When did it arrive in Boston?

Jeff earned his pilot's license and flew to visit his brother 200 miles away. A headwind of 20 mph slowed down the plane's speed

lora16 [44]

<h2>Plane's speed without wind i s 117.68 mph</h2>

Step-by-step explanation:

We have speed of plane without wind is x.

Distance to brothers place = 200 miles.

A headwind of 20 mph slowed down the plane's speed on the first leg of the trip

Speed to brothers place = x - 20

We have

Distance = Speed x Time

$200=(x-20)\times t_1\\\\t_1=\frac{200}{x-20}$

A tailwind of 20 mph sped up the plane on the return trip

Speed of return trip = x + 20

We have

Distance = Speed x Time

$200=(x+20)\times t_2\\\\t_2=\frac{200}{x+20}$

The entire trip took 3.5 hours.

That is

$t_1+t_2=3.5\\\\\frac{200}{x-20}+\frac{200}{x+20}=3.5\\\\200x+4000+2000x-4000=3.5(x-20)(x+20)\\\\400x=3.5x^2-1400\\\\7x^2-800x-2800=0\\\\\texttt{x = 117.68mph or x =-3.40mph}$

Plane's speed without wind i s 117.68 mph

7 0

2 years ago

One way to measure whether the trees in the Wade Tract are uniformly distributed is to examine the average location in the north

tatyana61 [14]

Answer:

Null hypothesis be H₀ : μ = 100

Alternative hypothesis Hₐ : μ < 100

There is sufficient statistical evidence to suggest that the average location of Wade Tract is 100

Step-by-step explanation:

Here we have;

Let our null hypothesis be H₀ : μ = 100 Average location of trees in the Wade Tract is 100

Our alternative hypothesis becomes Hₐ : μ < 100 at 95% confidence level

Proposed average location in Wade Tract, μ = 100

Sample mean, $\bar x$ = 99.74

Standard deviation, s = 58

Sample size, n = 584

The t test formula is therefore;

$t=\frac{\bar{x}-\mu }{\frac{s }{\sqrt{n}}}$

Therefore, with df = 584 -1 = 583, and α = (1 - 0.95)/2 = 0.025

We have $t_{\alpha /2}$ = -1.65

Plugging in the values into the t test formula, we have t = -0.108338, from which the p-value is given as p = 0.4569 which is much more than the value of α, therefore, we accept the null hypothesis as follows;

There is sufficient statistical evidence to suggest that the average location of Wade Tract = 100.

7 0

2 years ago

6.In a sample of 131 women with cosmetic dermatitis from using eye shadow, 12 were diagnosed with a nickel allergy. In a sample

aivan3 [116]

Answer:

a) $0.0916 - 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.0422$

$0.0916 + 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.1410$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using eye shadow at 95% of confidence is between (0.0422 and 0.1410)

b) $\hat p = \frac{25}{250}= 0.1$

$0.1 - 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.0628$

$0.1 + 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.1372$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using mascara at 95% of confidence is between (0.0628 and 0.1372)

c) For this case we see that both confidence intervals contains the value of 0.12 so then we can't conclude that only one group is referenced at the significance level of 0.05 used.

Step-by-step explanation:

Part a

The estimated proportion of women with cosmetic dermatitis from using eye shadow is given by:

$\hat p =\frac{12}{131}= 0.0916$

The confidence interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the proportion is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing we got:

$0.0916 - 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.0422$

$0.0916 + 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.1410$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using eye shadow at 95% of confidence is between (0.0422 and 0.1410)

Part b: A 95% confidence interval for the women with cosmetic dermatitis from using mascara

$\hat p = \frac{25}{250}= 0.1$

$0.1 - 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.0628$

$0.1 + 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.1372$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using mascara at 95% of confidence is between (0.0628 and 0.1372)

Part c: Suppose you are informed that the true proportion with a nickel allergy for one of the two groups (eye shadow or mascara) is .12. Can you determine which group is referenced? Explain.

For this case we see that both confidence intervals contains the value of 0.12 so then we can't conclude that only one group is referenced at the significance level of 0.05 used.

7 0

2 years ago

over the course of numbering every page in a book, a mechanical stamp printed 2929 individual digits. How many pages does the bo

Kazeer [188]

1. Page 1-9 (1digit/page x 9page) = 9 digits = 9 page

2. Page 10-99 (2digits/page x 90page)= 180 digits = 90 page

3. Page 100-999 (3digits/page x 900page)= 2700 digits = 900 page

2,929 total digits – (9 digits + 180 digits + 2700 digits)= 40 digits

Since there are only 4 digits after page 999 we divide 40/4 = 10, thus there are 10 more pages after 999

So the book has 1009 pages.

Hope this answer will be a good help for you.

4 0

2 years ago

The function f(x) = x2 is translated 7 units to the left and 3 units down to form the function g(x). Which represents g(x)? g(x)

Verdich [7]

Answer:

g(x)=(x+7)^2-3

Step-by-step explanation:

Given:

f(x)= x^2

Now we have to translate f(x) 7 units to the left and 3 units down to form the function g(x).

As per the rules of translation

when any parent function, in given case f(x)=x^2, is translated to 'a' units to the left then 'a' is added to the value of x. thus making f(x+a)

Also when the parent function is translated any 'a' units down then 'a' is subtracted from the value of function. thus making f(x)-a

Translating f(x), 7 units to the left

f(x+7)= (x+7)^2

Translating f(x+7), 3 units down

f(x+7)-3 = (x+7)^2-3

Hence new function g(x)=(x+7)^2-3!

7 0

2 years ago