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2 years ago

Four numbers have a mean of 24 Three of the numbers are 22,36 and 9 What is the other number? PLS HELP ME!

Mathematics

1 answer:

jolli1 [7]2 years ago

3 0

Answer:

Step-by-step explanation:

mean = add up everything and divide by the number of numbers there are.

so there are 4 numbers

x/4=24

3 of the numbers are 22, 36 , and 9.

So, (22+36+9+x) / 4 = 24

Multiply 4 by both sides

22+36+9+x=144

x=77.

Hope this helps!

#TeamRainbows!

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What is the value of y in the sequence below? 2,y,18, -54,162,

snow_lady [41]

First, let's check if the sequence is geometric or arithmetric.

If arithmetric, the sequence will have common difference.

Arithmetric

$\displaystyle \large{a_{n + 1} - a_n = d}$

d stands for a common difference. Common Difference means that sequences must have same difference after subtracting.

Geometric

$\displaystyle \large{ \frac{a_{n + 1}}{a_n} = r}$

r stands for a common ratio.

To find the value of y, you can check the sequence. If we try subtracting the sequences, the differences are different. That means the sequences are not arithmetric. That only leaves the geometric sequence.

Let's check by dividing sequences.

We have:

2,y,18,-54,162,...

Let's check by divide -54 by 18 and 162 by -54. We need to divide more than one so we can prove that the sequence is geometric.

$\displaystyle \large{ \frac{ - 54}{18} = - 3 } \\ \displaystyle \large{ \frac{ 162}{ - 54} = - 3}$

Hence, the sequence is geometric.

Because the common ratio is -3. Let these be the following:

$\displaystyle \large{ a_{n + 1} = y } \\ \displaystyle \large{ a_n = 2 } \\ \displaystyle \large{ r = - 3 }$

From the:

$\displaystyle \large{ \frac{a_{n + 1}}{a_n} = r}$

Substitute the values in.

$\displaystyle \large{ \frac{y}{2} = - 3}$

Multiply the whole equation by 2 to isolate y.

$\displaystyle \large{ \frac{y}{2} \times 2 = - 3 \times 2} \\ \displaystyle \large{ y = - 6}$

Therefore, the value of y is -6.

7 0

2 years ago

Every day your friend commutes to school on the subway at 9 AM. If the subway is on time, she will stop for a $3 coffee on the w

Shtirlitz [24]

Answer:

1.02% probability of spending 0 dollars on coffee over the course of a five day week

7.68% probability of spending 3 dollars on coffee over the course of a five day week

23.04% probability of spending 6 dollars on coffee over the course of a five day week

34.56% probability of spending 9 dollars on coffee over the course of a five day week

25.92% probability of spending 12 dollars on coffee over the course of a five day week

7.78% probability of spending 12 dollars on coffee over the course of a five day week

Step-by-step explanation:

For each day, there are only two possible outcomes. Either the subway is on time, or it is not. Each day, the probability of the train being on time is independent from other days. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

The probability that the subway is delayed is 40%. 100-40 = 60% of the train being on time, so $p = 0.6$

The week has 5 days, so $n = 5$

She spends 3 dollars on coffee each day the train is on time.

Probabability that she spends 0 dollars on coffee:

This is the probability of the train being late all 5 days, so it is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.6)^{0}.(0.4)^{5} = 0.0102$

1.02% probability of spending 0 dollars on coffee over the course of a five day week

Probabability that she spends 3 dollars on coffee:

This is the probability of the train being late for 4 days and on time for 1, so it is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{5,1}.(0.6)^{1}.(0.4)^{4} = 0.0768$

7.68% probability of spending 3 dollars on coffee over the course of a five day week

Probabability that she spends 6 dollars on coffee:

This is the probability of the train being late for 3 days and on time for 2, so it is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{5,2}.(0.6)^{2}.(0.4)^{3} = 0.2304$

23.04% probability of spending 6 dollars on coffee over the course of a five day week

Probabability that she spends 9 dollars on coffee:

This is the probability of the train being late for 2 days and on time for 3, so it is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{5,3}.(0.6)^{3}.(0.4)^{2} = 0.3456$

34.56% probability of spending 9 dollars on coffee over the course of a five day week

Probabability that she spends 12 dollars on coffee:

This is the probability of the train being late for 1 day and on time for 4, so it is P(X = 4).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{5,4}.(0.6)^{4}.(0.4)^{1} = 0.2592$

25.92% probability of spending 12 dollars on coffee over the course of a five day week

Probabability that she spends 15 dollars on coffee:

Probability that the subway is on time all days of the week, so $P(X = 5)$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{5,5}.(0.6)^{5}.(0.4)^{0} = 0.0778$

7.78% probability of spending 12 dollars on coffee over the course of a five day week

8 0

3 years ago

Establish for which constants c the following function is a density.

Valentin [98]

By "density" I assume you mean "probability density function". For this to be the case for $f(x)$ , we require

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=1$

Since

$f(x)=\begin{cases}cx^{1/2}&\text{for }0$

you have

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=\int_0^2cx^{1/2}\,\mathrm dx=\dfrac{2c}3x^{3/2}\bigg|_{x=0}^{x=2}=\dfrac{2^{5/2}c}3=1$

which means

$c=\dfrac3{2^{5/2}}=\dfrac3{4\sqrt2}$