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3 years ago

The graph shows the results of a race between Damon and Alex. Both boys begin the race at the same time.

Mathematics

2 answers:

Zarrin [17]3 years ago

4 0

Answer:

1st and 3rd statements are true.

Sep by step explanation:

We have been given a graph that shows the results of a race between Damon and Alex. Both boys begin the race at the same time.

Let us see which of the given option are correct.

1. Let us find Alex's speed using slope of the graph.

$\text{Slope}=\frac{80-0}{0-12}$

$\text{Slope}=\frac{80}{-12}$

$\text{Slope}=-6\frac{8}{12}$

$\text{Slope}=-6\frac{2}{3}$

Slope is showing remaining distance per second. The distance is decreasing $6\frac{2}{3}$ yards per second. Therefore, Alex is running at a rate of $6\frac{2}{3}$ yards per second and 1st statement is the true.

2. We can see from our graph that before starting the race Damon's distance from finish line is 100 yards and Alex's distance from finish line is 80 yards, therefore, Alex got a 20 yard head start.

Therefore, 2nd statement is false.

3. Let us find speed of Damon using slope.

$\text{Slope}=\frac{100-0}{0-10}$

$\text{Slope}=\frac{100}{-10}$

$\text{Slope}=-10$

We can see that slope is showing remaining distance per second. The distance is decreasing 10 yards per second therefore, Damon is running at a rate of 10 yards per second and 3rd statement is true.

4. We have already seen that Alex is given a 20 yard head start but he lost the race as he completed the race 2 seconds after Damon, therefore, 4th statement is false.

5. We can see from our graph that Alex completed race in 12 seconds, while Damon completed race in 10 seconds that is 2 minutes before Alex, therefore, 5th statement is false.

6. We have already seen that Damon runs at 10 yards per second and Alex runs at rate of $6\frac{2}{3}$ yards per second, therefore, last option is incorrect.

NeX [460]3 years ago

3 0

The first one, the 3rd one and the 5th one are true.

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3 years ago

Suppose approximately 75% of all marketing personnel are extroverts, whereas about 70% of all computer programmers are introvert

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Answer:

$P(x \ge 5) = 1.000$ ---- At least 5 from marketing departments are extroverts

$P(x=15) = 0.013$ ---- All from marketing departments are extroverts

$P(x = 0) = 0.002$ ---------- None from computer programmers are introverts

Step-by-step explanation:

See comment for complete question

The question is an illustration of binomial probability where

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$(a):\ P(x \ge 5)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

Using the complement rule, we have:

$P(x \ge 5) = 1 - P(x < 5)$

So, we have:

$P(x < 5) =P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)$

$P(x = 0) = ^{15}C_{0} * (75\%)^0 * (1 - 75\%)^{15 - 0} = 1 * 1 * (0.25)^{15} = 9.31 * 10^{-10}$

$P(x = 1) = ^{15}C_{1} * (75\%)^1 * (1 - 75\%)^{15 - 1} = 15* (0.75)^1 * (0.25)^{14} = 4.19 * 10^{-8}$

$P(x = 2) = ^{15}C_{2} * (75\%)^2 * (1 - 75\%)^{15 - 2} = 105* (0.75)^2 * (0.25)^{13} = 8.80 * 10^{-7}$

$P(x = 3) = ^{15}C_{3} * (75\%)^3 * (1 - 75\%)^{15 - 3} = 455* (0.75)^2 * (0.25)^{12} = 0.0000153$

$P(x = 4) = ^{15}C_{4} * (75\%)^4 * (1 - 75\%)^{15 - 4} = 1365 * (0.75)^4 * (0.25)^{11} = 0.000103$

So, we have:

$P(x < 5) = (9.31 * 10^{-10}) + (4.19 * 10^{-8}) + (8.80 * 10^{-7}) + 0.0000153 + 0.000103$

$P(x < 5) = 0.00011922283$

Recall that:

$P(x \ge 5) = 1 - P(x < 5)$

$P(x \ge 5) = 1 - 0.00011922283$

$P(x \ge 5) = 0.9998$

$P(x \ge 5) = 1.000$ --- approximated

$(b)\ P(x = 15)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x=15) = ^{15}C_{15} * 0.75^{15} * (1 - 0.75)^{15-15}$

$P(x=15) = 1 * 0.75^{15} * (0.25)^{0$

$P(x=15) = 0.013$

$(c)\ P(x = 0)$

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$p = 70\%$ --- proportion that are introverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x = 0) = ^{5}C_0 * (70\%)^0 * (1 - 70\%)^{5-0}$

$P(x = 0) = 1 * 1 * (0.30)^5$

$P(x = 0) = 0.002$

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Answer:

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Step-by-step explanation:

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3). LA - [One leg and one angle must be equal]

4). ASA - [Two angles and the side containing these angles should be equal]

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3 years ago

Read 2 more answers