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finlep [7]
1 year ago
8

Two cyclists , A and B , are going on a bike ride and are meeting at an orchard . They left home at the same time . Functions A

and B give their distance from the orchard , in miles , after riding for x hours . The functions are defined by these equations
Mathematics
1 answer:
vova2212 [387]1 year ago
5 0

Answer:

Cyclist A lives farther from the Orchard

Step-by-step explanation:

Given

A(x) =48.5 -21x

B(x) =42 -16.8x

Required:

Who lives farther from the Orchard.

To do this, we simply calculate the y intercept.

This represents the total distance traveled to get to the Orchard

To calculate y intercept, we set x = 0

So:

A(x) =48.5 -21x

A(0) = 48.5-21*0 =48.5-0 =48.5

B(x) =42 -16.8x

B(0) = 42 - 16.8*0 =42-0 =42

From the above computation:

A(0) > B(0) \to 48.5 > 42

Hence;

Cyclist A lives farther from the Orchard

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Calculate the side lengths a and b to two decimal places
Talja [164]

Answer:

The answer is (D) ⇒ a = 11.71 , b = 15.56

Step-by-step explanation:

* In ΔABC

∵ m∠A = 45°

∵ m∠B = 110°

∴ m∠C = 180 - 45 - 110 = 25°

By using the sin Rule

∵ a/sin(A) = b/sin(B) = c/sin(C)

∵ c = 7

∴ a/sin(45) = b/sin(110) = 7/sin(25)

∴ a = (7 × sin(45)) ÷ sin(25) = 11.71

∴ b = (7 × sin(110)) ÷ sin(25) = 15.56

∴ The answer is (D)

5 0
2 years ago
A tank contains 8000 L of pure water. Brine that contains 35 g of salt per liter of water is pumped into the tank at a rate of 2
OleMash [197]

Answer:

The concentration of salt in the tank approaches 35 \mathrm{g} / \mathrm{L},

Step-by-step explanation:

Data provide in the question:

Water contained in the tank = 8000 L

Salt per litre contained in Brine = 35 g/L

Rate of pumping water into the tank = 25 L/min

Concentration of salt \lim _{t \rightarrow \infty} C(t)=\lim _{t \rightarrow \infty} \frac{35 t}{320+t}

Now,

Dividing both numerator and denominator by t, we have

\lim _{t \rightarrow \infty} \frac{\frac{1}{t} 35 t}{\frac{1}{t}(320+t)}=\lim _{t \rightarrow \infty} \frac{35}{\frac{320}{t}+1}=35

Here,

The concentration of salt in the tank approaches 35 \mathrm{g} / \mathrm{L},

3 0
2 years ago
A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the
lbvjy [14]

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

Step-by-step explanation:

(a)

Let <em>X</em> = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

<em>X</em> = 610

<em>n</em> = 1000

Confidence level = 90%

The (1 - <em>α</em>)% confidence interval for population proportion is:

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

The sample proportion is:

\hat p=\frac{X}{n}=\frac{610}{1000}=0.61

The critical value of <em>z</em> for a 90% confidence level is:

z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

<u>Interpretation</u>:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let <em>X</em> = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

<em>X</em> = 440

<em>n</em> = 1000

Confidence level = 95%

The (1 - <em>α</em>)% confidence interval for population proportion is:

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

The sample proportion is:

\hat p=\frac{X}{n}=\frac{440}{1000}=0.44

The critical value of <em>z</em> for a 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend  is (0.42, 0.46).

<u>Interpretation</u>:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

  1. The confidence level
  2. Sample size
  3. Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of <em>z</em> in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

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Among the licensed drivers in the same age group, what is the probability that
vova2212 [387]

Answer:8%

Step-by-step explanation:

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