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4 years ago

WHAS UP MY PEEPS. I haven't been here in a year

Mathematics

2 answers:

AVprozaik [17]4 years ago

4 0

Answer:

cool

Step-by-step explanation:

Tems11 [23]4 years ago

4 0

Answer:

Nothing much. Been holding down the fort while u was gone

Step-by-step explanation:

You might be interested in

The Johnson twins were born nine years after their older sister. This year, the product of the three siblings ages is exactly 18

KATRIN_1 [288]

Answer: The age of the twins is 10 years.

Step-by-step explanation:

Suppose the age of older sister be x

Let the age of twins sister be x-9, as they are born after 9 years.

Product of their ages be

$x(x-9)(x-9)\\\\=x(x^2-18x+81)\\\\=x^3-18x^2+81x$

Sum of their ages will be

$x+x-9+x-9\\\\=3x-18$

According to question,

$1861+3x-18=x^3-18x^2+81x\\\\x^3-18x^2+81x-3x+18-1861=0\\\\x^3-18x^2+78x-1843=0\\\\x=19$

So, Age of older sister will be 19 years.

Age of the twins will be

$19-9=10\ years$

Hence, the age of the twins is 10 years.

3 0

3 years ago

If A || B and B | y, then ?

aniked [119]

Answere: I believe that the answere is C.

Step-by-step explanation:

Well,since the lines A and B are paralel and the line y is not paralel with any of then y and A are not paralel.Plus there is not a line called x in this particular equazion.If you have any questions , please contact me.

Yours sincerely,

Manos

8 0

3 years ago

In inches, Micah is 5 inches shorter than twice the height of Jasmine, who is 32 inches tall. How tall is Micah in inches?

Monica [59]

Micah is 5 inches shorter than twice the height of Jasmine

Micah = 5 inches shorter than 2 x Jasmine

Micah = 2 x Jasmine - 5 <em>(jasmine is 32 inches tall. Replace jasmine with 32)
</em>
Micah = 2 x 32 - 5

Micah = 64 - 5

Micah = 59

Micah is 59 inches tall

5 0

3 years ago

Read 2 more answers

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

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

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

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

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

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago

Micah’s bill for dinner was $35.50. He left the server an 18% tip. What is the total amount that Micah paid for dinner including

Lilit [14]

Answer:

$41.89

Step-by-step explanation:

Can I get brainliest please?

7 0

3 years ago

Read 2 more answers