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4 years ago

What is the area of triangle UVW

Mathematics

1 answer:

kogti [31]4 years ago

7 0

This is a polygon with vertices on the lattice. Let's use Pick's Theorem,

A = (1/2) B + I - 1

where A is the area, B is the number of lattice points on the boundary and I is the number of lattice points in the interior.

In addition to the 3 vertices there are 3 more boundary points on UV and 6 more on WV, none on UV, B=3+3+6=12. In the interior I count I=9 lattice points.

A = (1/2) 12 + 9 - 1 = 14

Answer: 14

Obviously they just want us to say this is a right triangle, so the legs are altitude and base,

A = (1/2) b h (1/2) |UW| |WV| = (1/2) (4) (7) = 14

That checks.

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Answer:

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Step-by-step explanation:

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Write the numerical expression in words then solve (98-78)×11

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4 years ago

The probability that a randomly selected teenager watched a rented video at least once during a week was 0.75. What is the proba

Whitepunk [10]

Answer:

There is a 75.65% probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week.

Step-by-step explanation:

For each teenager, there are only two possible outcomes. Either they watched a rented video at least once during a week, or they did not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

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

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

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 p is the probability of X happening.

In this problem we have that:

The probability that a randomly selected teenager watched a rented video at least once during a week was 0.75. This means that $p = 0.75$ .

What is the probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week?

Group of 7, so $n = 7$ .

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)$ .

In which

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

$P(X = 5) = C_{7,5}.(0.75)^{5}.(0.25)^{2} = 0.3115$

$P(X = 6) = C_{7,6}.(0.75)^{6}.(0.25)^{1} = 0.3115$

$P(X = 7) = C_{7,7}.(0.75)^{7}.(0.25)^{0} = 0.1335$

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) = 0.3115 + 0.3115 + 0.1335 = 0.7565$ .

There is a 75.65% probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week.

3 0

3 years ago

A manufacturer of pharmaceutical products analyzes specimens from each batch of a product to verify the concentration of the act

Alisiya [41]

Answer:

$0.82-2.977\frac{0.048}{\sqrt{15}}=0.783$

$0.82+2.977\frac{0.048}{\sqrt{15}}=0.857$

So on this case the 99% confidence interval would be given by (0.783;0.857)

So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X= 0.82$ represent the sample mean

$\mu$ population mean (variable of interest)

s=0.048 represent the sample standard deviation

n=15 represent the sample size

Solution to the problem

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

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

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=15-1=14$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that $t_{\alpha/2}=2.977$

Now we have everything in order to replace into formula (1):

$0.82-2.977\frac{0.048}{\sqrt{15}}=0.783$

$0.82+2.977\frac{0.048}{\sqrt{15}}=0.857$

So on this case the 99% confidence interval would be given by (0.783;0.857)

So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval

5 0

3 years ago