If there are 4 prizes of every 9 boxes how many prizes are in 54 boxes

Ryan works at at the donut shop where he makes $10.25 per hour . He also works part time at the school bookstore where he makes

SCORPION-xisa [38]

Answer:

6 hours

Step-by-step explanation:

Let

x----> the number of hours worked at the donut shop

y----> the number of hours worked at the school bookstore

we know that

x+y=20

x=20-y -----> equation A

10.25x+8.75y=196 -----> equation B

Substitute equation A in equation B and solve for y

10.25(20-y)+8.75y=196

205-10.25y+8.75y=196

10.25y-8.75y=205-196

1.5y=9

y=6 hours at the school bookstore

7 0

3 years ago

Quantos que é -2 ao quadrado

Olegator [25]

No estoy segura e -2 ao quadrado esa es toda la pregunta

5 0

3 years ago

Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1

svet-max [94.6K]

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

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 $s = \frac{\sigma}{\sqrt{n}}$ .