What is ſ(3)? A. -9 B.-1 C.1 D.9

A. If you have 8 bags, each weighing 50 lbs., and have to weigh

Elza [17]

No, you do not have enough

8 0

3 years ago

free shopping with her mother they pay $2 per pound for tomatoes at the vegetable stand find the constant proportionality in thi

nlexa [21]

Equation would be y=2x

So how you would use it is let’s say they bought 10
So y=2(10) so 20 dollars for 10 pounds

5 0

4 years ago

Read 2 more answers

PLEASE HELP IM SO CONFUSED!!!

Step2247 [10]

Problem 1

The median weight of the Welsh corgis is 27 pounds. This is found in the table in row 1, column 2. Recall that the median is the middle most value, so that means 50% of the Welsh corgi weights will be less than 27 pounds, and also 50% of the weights are greater than 27 pounds.

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Problem 2

The Welsh corgis have the greater variation in weights. Why is this? There are two pieces of evidence:

The first of which is that the range for the first row (7) is larger than the range in the second row (2). The range is the difference between the max and min. The larger the range, the more spread out the data set will be.

The second piece of evidence is that the MAD (mean absolute deviation) for the Welsh corgis is larger compared to the MAD for the cocker spaniels group. The MAD is another way to measure the spread of a data set. Like the range, the larger the MAD gets, the more spread out the data is. It measures how far on average any given value is from the mean. It's similar to the standard deviation and variance, which are also measures of spread.

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Problem 3

Both dog breeds have the same mean value, and roughly the same median. Both the mean and median are measures of center. Based on just these two pieces of info alone, we could easily conclude that both dog groups have the same distribution more or less. Furthermore, we could conclude that both groups have roughly the same max weight value.

However, we discussed in problem 2 that the first group (Welsh corgi) is more spread out, so it's likely that could heavily imply that the min and max values are further from the center compared to the second group of dogs. Having a larger max would then mean that the corgi group has the largest weight.

In short, the answer is "Welsh corgi" due to this group having a greater spread, and that pushes the max further away from the center.

7 0

3 years ago

Read 2 more answers

According to an airline, flights on a certain route are on time % of the time. Suppose flights are randomly selected and the n

neonofarm [45]

Answer:

(a) Explained below.

(b) 0.0294

(c) 0.0173

(d) 0.09827

(e) 0.0452

Step-by-step explanation:

The complete question is:

According to an airline, flights on a certain route are on time 80% of the time. Suppose 25 flights are randomly selected and the number of on-time flights is recorded.

(a) Explain why this is a binomial experiment.

(b) Find and interpret the probability that exactly 16 flights are on time.

(c) Find and interpret the probability that fewer than 16 flights are on time.

(d) Find and interpret the probability that at least 16 flights are on time.

(e) Find and interpret the probability that between 14 and 16 flights, inclusive, are on time.

Solution:

(a)

Let the random variable X be defined as the number of on-time flights.

A Binomial experiment has the following properties:

There are a fixed number of trials (n).
Each trial are independent of the others.
Each trial has only two outcomes: Success and Failure
Each trial has the same probability of success (p).

If a random variable X is used in an experiment and the experiment has all the above mentioned properties, then the random variable X is known as a binomial random variable.

All of these properties can be confirmed for the random variable X.

Thus, this is a binomial experiment.

(b)

Compute the probability that exactly 16 flights are on time as follows:

$P(X=16)={25\choose 16}(0.80)^{16}(0.20)^{25-16}$

$=2042975\times 0.0281475\times 0.000000512\\=0.029442375072\\\approx 0.0294$

Thus, the probability that exactly 16 flights are on time is 0.0294.

(c)

Compute the probability that fewer than 16 flights are on time as follows:

$P(X$

$=0.0000+0.0000+....+0.011777\\=0.0173$

Thus, the probability that fewer than 16 flights are on time is 0.0173.

(d)

Compute the probability that at least 16 flights are on time as follows:

$P(X\geq 16)=1-P(X$

$=1-0.0173\\=0.9827$

(e)

Compute the probability that between 14 and 16 flights, inclusive, are on time as follows:

$P(14\leq X\leq 16)=\sum\limits^{16}_{x=14}{{25\choose x}(0.80)^{x}(0.20)^{25-x}}$

$=0.004+0.0118+0.0294\\=0.0452$

Thus, the probability that between 14 and 16 flights, inclusive, are on time is 0.0452.

8 0

3 years ago

A triangle has sides with lengths of 5x-7,3x-4 and 2x-6. What is the perimeter of the triangle?

boyakko [2]

The perimeter is the sum of all the sides, so P = (5x -7) + (3x-4) + (2x -6). Then group like terms so you have p = 5x + 3x + 2x -7-4-6 and end up with P = 10x - 17

7 0

3 years ago