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3 years ago

X subtract 9 equals 3 / 5X

Mathematics

1 answer:

Korvikt [17]3 years ago

7 0

Answer:

$x=22\frac{1}{2}=22.5$

Step-by-step explanation:

$x-9=\frac{3}{5}x$

First, lets get rid of the fraction. I did this by multiplying both sides by $5$ .

$5(x-9)=5(\frac{3}{5}x)\\5x-45=\frac{5}{1} *\frac{3}{5}x\\5x-45=\frac{15}{5}x\\5x-45=3x$

We want to isolate $x$ on one side of this equation. Let's put any values with $x$ on one side of the equation, and normal integers on the other.

$2x=45$

Divide both sides by $2$ .

$x=\frac{45}{2}$

If your teacher wants you to leave your final answer as an improper fraction, your final answer is this. If they want it to be a mixed number or decimal, your final answer will be:

$x=22\frac{1}{2}=22.5$

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To avoid detection at customs, a traveler has placed 6 narcotic tablets in an Urn containing 9 vitamin pills that are similar in

Alexus [3.1K]

Answer:

$\frac{4}{91}$

Step-by-step explanation:

We are told that to avoid detection at customs, a traveler has placed 6 narcotic tablets in an Urn containing 9 vitamin pills that are similar in appearance.

To find the probability that the traveler will be arrested for illegal possession of narcotics we will use theoretical probability formula.

$\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}$

Favorable outcomes will be $_{3}^{6}\textrm{C}$ .

$\text{Favorable outcomes}=_{3}^{6}\textrm{C}=\frac{6!}{(6-3)!3!}$

$\text{Favorable outcomes}=\frac{6!}{3!3!}$

$\text{Favorable outcomes}=\frac{6\cdot 5\cdot 4\cdot3!}{3\cdot 2\cdot 1\cdot3!}$

$\text{Favorable outcomes}=\frac{6\cdot 5\cdot 4}{3\cdot 2}$

$\text{Favorable outcomes}=5\cdot 4=20$

Now let us find number of possible outcomes.

$\text{Number of possible outcomes}=_{3}^{15}\textrm{C}=\frac{15!}{(15-3)!3!}$

$\text{Number of possible outcomes}=\frac{15!}{12!3!}$

$\text{Number of possible outcomes}=\frac{15\cdot 14\cdot 13\cdot 12!}{12!\cdot 3\cdot 2\cdot 1}$

$\text{Number of possible outcomes}=\frac{15\cdot 14\cdot 13}{ 3\cdot 2\cdot 1}$

$\text{Number of possible outcomes}=5\cdot 7\cdot 13=455$

Upon substituting our values in probability formula we will get,

$\text{Probability that the traveler will be arrested for illegal possession of narcotics}=\frac{20}{455}$

Upon dividing numerator and denominator by 5 we will get,

$\text{Probability that the traveler will be arrested for illegal possession of narcotics}=\frac{4}{91}$

Therefore, the probability that the traveler will be arrested for illegal possession of narcotics is $\frac{4}{91}$ .

6 0

3 years ago

What property is being used in 2a+5a=(2+5)a

Tems11 [23]

Answer:

distributive property

5 0

3 years ago

Which ordered paired IS NOT a solution of y = 3x - 8

vazorg [7]

2,-2 seems to be the answer

8 0

3 years ago

nancy spent 7/8 hour working out at the gym she spent 5/7 of that time lifting weights what fraction of an hour did sje spentld

Juli2301 [7.4K]

$\frac{9}{56}\\$ Fraction of hour was spent by NAncy in lifting weights

Step-by-step explanation:

Time spent at the gym by Nancy = $\frac{7}{8}\,hour$

Time spent at lifting weights = $\frac{5}{7}\,hour$

What fraction of hour she spent in lifting weights?

Solving:

Fraction of hour she spent in lifting weights= Time spend at the gym-Time spent at lifting weights

Fraction of hour she spent in lifting weights= $\frac{7}{8}\,hour-\frac{5}{7}\,hour$

$=\frac{49}{56}-\frac{40}{56}\\=\frac{49-40}{56}\\=\frac{9}{56}\\$

So, $\frac{9}{56}\\$ Fraction of hour was spent by Nancy in lifting weights

Keywords: Word Problems involving Fractions

Learn more about Word Problems involving Fractions at:

#learnwithBrainly

7 0

3 years ago

The table below shows data from a survey about the amount of time students spend doing homework each week. The students were in

nadezda [96]

Answer:

The correct option is;

Both spreads are best described by the standard deviation

Step-by-step explanation:

The given information are;

, College High School

High, 20 20

Low, 6 3

Q₁, 8 5.5

Q₃, 18 16

IQR, 10 10.5

Median, 14 11

Mean, 13.3 11

σ, 5.2 5.4

Checking for outliers, we have

College

Q₁ - 1.5×IQR gives 8 - 1.5×10 = -7

Q₃ + 1.5×IQR gives 18 + 1.5×10 = 33

For high school

Q₁ - 1.5×IQR gives 5.5 - 1.5×10.5 = -10.25

Q₃ + 1.5×IQR gives 16 + 1.5×10.5 = 31.75

Therefore, there are no outliers and the data is representative of the population

From the data, for the college students, it is observed that the difference between the mean, 13.3 and Q₁, 8, and between Q₃, 18 and the mean,13.3 is approximately the standard deviation, σ, 5.2

The difference between the low and the high is also approximately 3 standard deviations

Therefore the college spread is best described by the standard deviation

Similarly for the high school students, the IQR is approximately two standard deviations, the difference between the mean, 11 and Q₁, 5.5, and between Q₃, 16 and the mean,11 is approximately the standard deviation, σ, 5.4

Therefore the high school spread is also best described by the standard deviation.

4 0

3 years ago