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Westkost [7]
2 years ago
5

Write an equation of a parabola that has two x-intercepts and a minimum vertex.

Mathematics
2 answers:
Gre4nikov [31]2 years ago
6 0
The answer is B I hope this helps !!
LuckyWell [14K]2 years ago
5 0
B is correct ok yesssss
You might be interested in
A ski lift is designed with a total load limit of 20,000 pounds. It claims a capacity of 100 persons. An expert in ski lifts thi
Yanka [14]

Answer:

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the sum of n values of a distribution, the mean is \mu \times n and the standard deviation is \sigma\sqrt{n}

An expert in ski lifts thinks that the weights of individuals using the lift have expected weight of 200 pounds and standard deviation of 30 pounds. 100 individuals.

This means that \mu = 200*100 = 20000, \sigma = 30\sqrt{100} = 300

If the expert is right, what is the probability that a random sample of 100 independent persons will cause an overload

Total load of more than 20,000 pounds, which is 1 subtracted by the pvalue of Z when X = 20000. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{20000 - 20000}{300}

Z = 0

Z = 0 has a pvalue of 0.5

1 - 0.5 = 0.5

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

5 0
2 years ago
Find the missing side length of the triangle 48 units 14 units​
Tpy6a [65]

Answer:

what are the options? I would say something like 50-60

5 0
3 years ago
Using the number line below to compare the list of numbers. Place the numbers in increasing order.
Ivanshal [37]

Answer:

-2/5

-1.5

.3

1 4/5

Step-by-step explanation:

The higher the negatice number, the lower it is. Hope this helps:)

4 0
2 years ago
Find the area of a triangle with vertices (1,5), (1, -2), (3,-2).
igomit [66]
The area is 7 units squared.

3-1 = 2
5--2 = 7
2x7 = 14
14/2 = 7
5 0
3 years ago
At a local University, it is reported that 81% of students own a wireless device. If 5 students are selected at random, what is
Sphinxa [80]

Answer:

34.87% probability that all 5 have a wireless device

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they own a wireless device, or they do not. The probability of a student owning a wireless device is independent from other students. So we use the binomial probability distribution to solve this question.

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 combinations 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.

81% of students own a wireless device.

This means that p = 0.81

If 5 students are selected at random, what is the probability that all 5 have a wireless device?

This is P(X = 5) when n = 5. So

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

P(X = 5) = C_{5,5}.(0.81)^{5}.(0.19)^{0} = 0.3487

34.87% probability that all 5 have a wireless device

5 0
3 years ago
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