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A point is chosen randomly on AD. What is the probability that the point is on BC?

Nadusha1986 [10]

I’m pretty sure it’s b but idk 100% sorry if it’s wrong

8 0

3 years ago

A store buys frozen burritos from a supplier for $1.40 each. The store adds a markup of 80% to determine the retail price. This

boyakko [2]

$1.40 - price from a supplier = 100%
100% + 80% = 180% = 1,8 - the retail price

1,40 * 1,8 = $2,52 - the retail price

100% - 25% = 75% = 0,75 - on sale for 25% off

2,52 * 0,75 = $1,89 - the sale price.

8 0

3 years ago

Is 92 1/2 pounds equivalent to 1,480 ounces

kow [346]

Hi there!

Since

16 ounces = 1 pound,

We should divide 1480 by 16 and see if it equals 92.5.

1480 ÷ 16 = 92.5
92.5 = 92.5

Answer:
Yes.

Hope this helps!

8 0

3 years ago

Read 2 more answers

If the level of significance of a hypothesis test is raised from 0.005 to 0.2, the probability of a type ii error will:_________

LiRa [457]

The probability of type II error will decrease if the level of significance of a hypothesis test is raised from 0.005 to 0.2.

<h3 /><h3>What is a type II error?</h3>

A type II error occurs when a false null hypothesis is not rejected or a true alternative hypothesis is mistakenly rejected.

It is denoted by 'β'. The power of the hypothesis is given by '1 - β'.

<h3>How the type II error is related to the significance level?</h3>

The relation between type II error and the significance level(α):

The higher values of significance level make it easier to reject the null hypothesis. So, the probability of type II error decreases.
The lower values of significance level make it fail to reject a false null hypothesis. So, the probability of type II error increases.
Thus, if the significance level increases, the type II error decreases and vice-versa.

From this, it is known that when the significance level of the given hypothesis test is raised from 0.005 to 0.2, the probability of type II error will decrease.

Learn more about type II error of a hypothesis test here:

brainly.com/question/15221256

#SPJ4

7 0

1 year ago

A website manager has noticed that during the evening hours, about 5 people per minute check out from their shopping cart and m

Over [174]

Answer:

a) Poisson distribution

b) 99.33% probability that in any one minute at least one purchase is made

c) 0.05% probability that seven people make a purchase in the next four minutes

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

5 people per minute check out from their shopping cart and make an online purchase.

This means that $\mu = 5$

a) What model might you suggest to model the number of purchases per minute? 

The only information that we have is the mean number of an event(purchases) in a time interval. Each event is also independent fro each other. So you should suggest the Poisson distribution to model the number of purchases per minute.

b) What is the probability that in any one minute at least one purchase is made? 

Either no purchases are made, or at least one is. The sum of the probabilities of these events is 1. So

$P(X = 0) + P(X \geq 1) = 1$