How do you Simply-6(150x^3y^7)

Help with homework please

BabaBlast [244]

Answer:

1

Step-by-step explanation:

Start at -0.5 and go toward 0.5.

From -0.5 to zero, it's a distance of 0.5.

From zero to 0.5, it's also a distance of 0.5.

Add the two distances, 0.5 + 0.5 = 1.

The distance between -0.5 to 0.5 is 1.

6 0

3 years ago

The weather report stated there was a 2/3 chance of rain on Friday, but it was more likely to rain on Saturday than Friday. What

Juliette [100K]

You don't have any choices here, so we can just find the range of the answer. Then, you can select the correct answer.

The probability must be greater than 66.67%: P > 66.67% (or 2/3).

The chance of rain on Friday is 2/3. If it is more likely on Saturday, then it must be more than 2/3. 2/3 is about 66.67% percent.

6 0

3 years ago

What is 10x5÷2 equal

Viefleur [7K]

$10 \times 5 = 50$
$50 \div 2 = 25$
$25$

5 0

2 years ago

Read 2 more answers

Suppose that the weights of airline passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation

Leto [7]

Answer:

There is a 24.51% probability that he weight of a bag will be greater than the maximum allowable weight of 50 pounds.

Step-by-step explanation:

Problems of normally distributed samples 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

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

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

Suppose that the weights of airline passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation of 3.09 pounds, so $\mu = 47.88, \sigma = 3.09$