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3 years ago

Two isosceles triangles are similar. The first

Mathematics

1 answer:

galina1969 [7]3 years ago

3 0

I think the answer may be 11

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Student scored in the 83rd

melamori03 [73]

Answer:b

Step-by-step explanation:

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3 years ago

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and tes

mart [117]

Answer:

44% is the probability

Step-by-step explanation:

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3 years ago

During which time period does Landon's elevation change the fastest? Explain how you know?

bogdanovich [222]

<h3>1. How many inches per minute does London's elevation change between 4 minutes and 8 minutes. </h3>

The question actually asks for the slope of the line that stands for the points $(4,3) \ and \ (8,6)$ why? because the questions tells us that London's elevation changes between 4 minutes and 8 minutes here. Hence, to find the slope of this line we have to use the following formula:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ But: \\ \\ P(x_{1},y_{1})=P(4,3) \\ P(x_{2},y_{2})=P(8,6)$

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ But: \\ \\ P(x_{1},y_{1})=P(4,3) \\ P(x_{2},y_{2})=P(8,6) \\ \\ So: \\ \\ m=\frac{6-3}{8-4}=0.75in/min$

<em>So London's elevation changes 0.75 inches per minute</em>

<h3>2. During which time period does London's elevation change the fastest?</h3>

The greater the absolute value of the slope of the line the faster London's elevation changes. Since this is a Piecewise function, we must analyze each period.

FIRST:

→ Between 0 minutes and 4 minutes the function is constant, so there is no any change here.

→ Between 10 minutes and 14 minutes the function is constant, so there is no any change here.

→ Between 18 minutes and 22 minutes the function is constant, so there is no any change here.

So the solution is not in these parts of the function.

SECOND:

→ Between 4 minutes and 10 minutes the function has a positive slope, so there is change here.

In the previous item we calculated the slope between 4 and 8 minutes that is the same slope between 4 and 8 minutes and equals 0.75.

→ Between 14 minutes and 18 minutes the function has a positive slope, so there is change here.

Let's take two points here, say, $(16,5) \ and \ (18,3)$

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ But: \\ \\ P(x_{1},y_{1})=P(16,5) \\ P(x_{2},y_{2})=P(18,3) \\ \\ So: \\ \\ m=\frac{3-5}{18-16}=-1 in/min$

As you can see, the absolute value here is 1 that is greater than 0.75.

<em>In conclusion, London's elevation changes the fastest between 14 and 18 minutes</em>

7 0

3 years ago

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago

Ben os travelling home to brimingham on the m1 and m42. both motorways have road works. He has a 50-50 chance of getting delay o

Contact [7]

The answer is 1/8 or 12.5%

8 0

3 years ago