Desiree went bowling. It was $4 to rent the shoes and $3.25 per game. If she spent a total of $30 how many games did she bowl?

What are the terms in the expression 350n-30n+350-(50+10n)

trapecia [35]

Terms- 350n, 30n, 350, and (50+10n) (four terms)
coefficients- 350,30,and 10

5 0

2 years ago

Please help!!! (With one or all the problems, anything is greatly appreciated if you understand probability)

professor190 [17]

Here are the answers to problem 1..

a) 11

b) 5/36

c) 25 Times because 7 has a probability of 16.667%

d) 25 Times because 10, 11, and 12 have a probability sum of 16.667% ironically...

Hope this helps. :)

6 0

3 years ago

WARNING : NON SENSE ANSWER REPORT TO MODERATOR

Rzqust [24]

Answer:

$\frac{180}{147}$

Step-by-step explanation:

Simplify $(\frac{3}{-7} - \frac{11}{21} )$

$=> \frac{(-3 \times 3) - 11}{21}$

$=> \frac{-9 - 11}{21}$

$=> \frac{-20}{21}$

Find the additive inverse of $\frac{-20}{21}$ by using its property - <em>"Sum of a number & its additive inverse is always zero". </em>Assume that 'x' is an additive inverse of $\frac{-20}{21}$ .

$=> x + \frac{-20}{21} = 0$

$=> x = 0 - (-\frac{20}{21}) = \frac{20}{21}$

Simplify $(\frac{9}{5} \div \frac{7}{5} )$

$=> \frac{9}{5} \times \frac{1}{\frac{7}{5} }$

$=> \frac{9}{5} \times \frac{5}{7}$

$=> \frac{9}{7}$

Now, find the product of $\frac{9}{7}$ & $\frac{20}{21}$

$=> \frac{9}{7} \times \frac{20}{21}$

$=> \frac{180}{147}$

3 0

3 years ago

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1262 and a s

Andrew [12]

Answer:

a) 1186

b) Between 1031 and 1493.

c) 160

Step-by-step explanation:

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean of 1262 and a standard deviation of 118.

This means that $\mu = 1262, \sigma = 118$