It costs $25 for every 2-hour art class, how much would it cost for 6 hours of art classes

Mathematics

2 answers:

e-lub [12.9K]3 years ago

7 0

Answer:

$75

Step-by-step explanation:

If it costs $25 for two hours and we want 6 hours, we simply multiply it by 3 because 6/2 = 3.

$25 * 3 = $75

valkas [14]3 years ago

7 0

Answer:

The answer $75

Step-by-step explanation:

you add $25 each 2 hours and it would be $75 simple

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4a/3-b/4=6<br> 5a/6+b=13<br> Solve linear system by elimination

kondor19780726 [428]

$\bf \cfrac{4a}{3}-\cfrac{b}{4}=6
\qquad \qquad 
\cfrac{5a}{6}+b=13
\\\\\\
\textit{let us remove the denominators off those folks}\\
\textit{by multiplying the first one by 12, both sides}\\
\textit{and the second one by 6, both sides, thus}
\\\\\\
12\left( \cfrac{4a}{3}-\cfrac{b}{4} \right)=12(6)\implies 16a-4b=72
\\\\\\
6\left( \cfrac{5a}{6}+b \right)=6(13)\implies 5a+6b=78$

$\bf \textit{now, let's do the elimination}
\\\\

\begin{array}{llll}
16a-4b=72&\boxed{\times 3}\implies &48a-\underline{12b}=216\\\\
5a+6b=78&\boxed{\times 2}\implies &10a+\underline{12b}=156\\
&&--------\\
&&58a+0\quad=372
\end{array}$

solve for "a", once you get "a", plug it back into either equation to get "b"

3 0

2 years ago

The lengths of nails produced in a factory are normally distributed with a mean of 4.84 centimeters and a standard deviation of

Helga [31]

Answer:

Top 3%: 4.934 cm

Bottom 3%: 4.746 cm

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 4.84, \sigma = 0.05$