In Manchester, 3 of the 18 radio stations play country music.

Mathematics

1 answer:

Kazeer [188]2 years ago

8 0

Answer:

There is a one in six chance that a randomly selected radio plays country music.

Step-by-step explanation:

There is a 3/18 chance but that can be simplified if you divide both sides by 3 to get 1/6

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An investment firm invested in two companies last year. They invested $11,000 in Company A and made a profit of 24% . They inves

Tema [17]

First off, "whatever%" of "anything" is just (whatever/100) * anything

part A)

so... firm A got 11,000 invested, turned a profit of 24%, how much is 24% of 11000? well (24/100) * 11000

firm B got 14,000 invested, and returned 15% in profits, how much is 15% of 14000? (15/100) * 14000

part B)

for the amounts above, we get 2640 and 2100 respectively

so, the total profit "amount" is 2640 + 2100 or 4740

the total investement was 11000+14000 = 25000

if 25000 is the 100%, how much is 4740 in percentage?

$\bf \begin{array}{ccllll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
25000&100\\
4740&x
\end{array}\implies \cfrac{25000}{4740}=\cfrac{100}{x}$

solve for "x"

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3 years ago

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-5/k=?/8k complete the expression

Nostrana [21]

?=-40 use -5/k)*(8k here u go

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3 years ago

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Suppose you pay $1.00 to roll a fair die with the understanding you will get $3.00 back rolling a 4 or a 2, and nothing otherwis

nikdorinn [45]

There are 2 chances out of 6 to win $3

Every 6 rolls you are expected to win $6

Every 3 rolls you are expected to win $3.

On average, even though it's impossible, every roll you would make $1.

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3 years ago

The U.S. Bureau of Labor Statistics reports that of persons who usually work full-time, the average number of hours worked per w

alukav5142 [94]

Answer:

The standard deviation of number of hours worked per week for these workers is 3.91.

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem we have that:

The average number of hours worked per week is 43.4, so $\mu = 43.4$ .

Suppose 12% of these workers work more than 48 hours. Based on this percentage, what is the standard deviation of number of hours worked per week for these workers.

This means that the Z score of $X = 48$ has a pvalue of 0.88. This is Z between 1.17 and 1.18. So we use $Z = 1.175$ .