Question 4 (1 point)

Mathematics

1 answer:

Aleksandr-060686 [28]3 years ago

5 0

2:3

6 ÷ 2 = $3 (cost of 1 pack of butter cookies)

3 x 3 = 9 (cost of 3 packs of butter cookies)

17 - 9 = 8 (cost of 4 packs of chocolate cookies)

8 ÷ 4 = 2 (cost of one pack of chocolate cookies)

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anne buys used cars at auction for 2000 per car. there is 150 fee to take part in the auction. If anne buys 13 used cars how muc

katrin [286]

How to understand the problem is like this:
Total Cost = Entrance Fee + Total Car cost
So we know that the Entrance Fee is $150...
Total Cost = $150 + Total Car cost
So we know that Anne bought 13 cars, each at $2,000...
Total Cost = $150 + ($2,000 x 13)
We multiply because each car is $2,000 and Anne bought 13 of them.
So it works out like this:
Total Cost = $150 + ($2,000 x 13)
Total Cost = $150 + (26000)
Total Cost = $26150

5 0

3 years ago

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st

Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

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

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so $\mu = 0.5, \sigma = 0.05$ .