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Lubov Fominskaja [6]
4 years ago
6

At a certain high school, 25 students play only stringed instruments, 15 students play only brass instruments, and 5 students pl

ay both. There are also 5 students who play neither stringed nor brass instruments. What is the probability that a randomly selected student plays both a stringed and brass instrument?
Mathematics
1 answer:
Oksana_A [137]4 years ago
3 0

Answer:

1/10

Step-by-step explanation:

First, let's find the total number of students (25+15+5+5). We don't need to worry about them overlapping, as the questions says "only." We get the total of 50.

Out of the 50 students, there are 5 students that play both instruments. This means that 5 is our numerator and 50 is our denominator, giving us 5/50 or 1/10.

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The weights of the apples grown on an orchard are normally distributed. The mean weight has been m0 = 9.500 ounces. Because of t
diamong [38]

Answer:

t(s) is in rejection zone then we reject H₀.

Bad weather indeed make apples weight decrease

Step-by-step explanation:

Normal Distribution

population mean    μ₀   = 9.5 ou

sample size   =  n   =  16  then we should apply t-student table

degree of fredom   df  =  n  - 1     df =  16  -  1    df  = 15

1.-Test  hypothesis

H₀     null  hypothesis                          μ₀   =  9.5    

Hₐ alternative hypothesis                   μ₀  <  9.5

One left tail-test

2.-Confidence level 95 %

α  =  0,05    and   df  = 15    from t-student table we get  t(c)  =  - 1.761

3.-Compute t(s)  

t(s)  =  [ μ  -  μ₀  ] /√s/n         t(s)  = (9.32  -  9.5 )* √16 / 0.18

t(s)  = -  0.18*√16 / 0.18

t(s)  = - 4

4.-Compare  t(s)   and t(c)

t(s) < t(c)     -4  <  - 1.761

Then  t(s)  is in the rejection zone.

5.- Decision

t(s) is in rejection zone then we reject H₀.

Farmer conclude that bad weather make apples weight decrease

6 0
4 years ago
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ch4aika [34]
It would be 100. Hope it helps!
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What’s 5/9 into a decimal and a percentage
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The answer to that problem is 0.55555556
4 0
3 years ago
Read 2 more answers
I’m about to cry from frustration. Help please before I die
kvv77 [185]

Answer:

\frac{1}{3}x+16=37

63 Jelly Beans

Step-by-step explanation:

The unknown is the number of jelly beans originally in the bag or x

First he had x jelly beans

The he ate one-third of them

\frac{1}{3}x

He then ate 16 more jelly beans

\frac{1}{3}x+16

This was equal to 37 jelly beans

\frac{1}{3}x+16=37

This is the equation

Now solve for x

Subtract 16 from both sides

\frac{1}{3}x=21

Multiply both sides by 3

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7 0
3 years ago
Read 2 more answers
I'm blanking on how to do this, I learned it so long ago, any help would be greatly appreciated. More interested on how to do it
anyanavicka [17]

Answer:

\dfrac{16 y^{22}}{x^{10}z^{10}}

Step-by-step explanation:

Given expression is ,

\sf\longrightarrow \bigg(\dfrac{2x^3y^{-5}z^8}{8x^{-2}y^6z^3}\bigg)^{-2}

This would be simplified using the law of exponents , some of which I will use here are ,

  • (an)^m = a^m n^m
  • \dfrac{a^m}{a^n}=a^{m-n}

  • a^m a^n = a^{m+n}
  • a^{-n} = \dfrac{1}{a^n}

Using the above laws ,

\sf \longrightarrow \bigg[ \dfrac{2}{8} \bigg(\dfrac{x^3}{x^{-2}}\bigg)\bigg(\dfrac{y^{-5}}{y^6}\bigg)\bigg(\dfrac{z^8}{z^3}\bigg)  \bigg]^{-2}

Using the second law mentioned above , we have,

\sf \longrightarrow \bigg[ \dfrac{1}{4}(x^{3+2})(y^{-5-6})(z^{8-3})\bigg]^{-2}

Simplify ,

\sf \longrightarrow \bigg[\dfrac{1}{4} x^5y^{-11}z^5\bigg]^{-2}

Using the first law mentioned above , we have,

\sf \longrightarrow \bigg[ \dfrac{1}{4^{-2}} x^{5(-2)} y^{-11(-2)} z^{5(-2)}\bigg]

Simplify,

\sf \longrightarrow 4^2x^{-10}y^{22} z^{-10}

Finally using the fourth law mentioned above , we have ,

\sf \longrightarrow \boxed{\bf \dfrac{16 y^{22}}{x^{10}z^{10}}}

<h3>Option K is the correct answer.</h3>
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2 years ago
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