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Elina [12.6K]
3 years ago
7

A school club sold 300 shirts. 31% were sold to fifth-graders, 52% were sold to sixth graders, and the rest were sold to teacher

s. How many shirts were sold to each group -- fifth graders, sixth graders, and teachers? Explain or show your reasoning.
Mathematics
1 answer:
Valentin [98]3 years ago
6 0

Answer:

Fifth graders= 93 shirts

Sixth graders= 156 shirts

Teachers= 51 shirts

Step-by-step explanation:

31% of 300 is 93

52% of 300 is 156

The total number of shirts sold to fifth graders and sixth graders is:

156+93= 249

The number of shirts sold to teachers is:

300-249= 51

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Lina20 [59]

Answer:

              t=\dfrac{k-rs}{3m}

Step-by-step explanation:

It's like solving any equation, except we have letters instead of numbers .

k = 3tm + rs           {subtract rs from both sides}

k - rs = 3tm            {divide both sides by (3m)}

\dfrac{k-rs}{3m}=t              {swap sides}

t=\dfrac{k-rs}{3m}            m≠0    

{m must be ≠0 because  you cannot divide by 0. There is no t for m=0}    

4 0
2 years ago
An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline
vivado [14]

Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

The success probability = 5 % = 0.05

Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have  a seat, we have  

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 =  5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

 z = \frac{x-np}{\sqrt{np(1-p)}  } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} } =  - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = \sqrt{npq} = 2.233

P (x≤100) = P(z = P(z<0.34) = 0.63307.

6 0
3 years ago
7968 divided by 22 estimating
Anvisha [2.4K]

Answer:

363

Step-by-step explanation:

7 0
2 years ago
2. In a relationship between variables, what is the variable called that changes in response to
valkas [14]

Answer:

A dependent variable

Step-by-step explanation:

is a variable whose value will change depending on the value of another variable, called the independent variable. Dependent variables are also known as outcome variables, left-hand-side variables, or response variables.

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natima [27]

 

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.

\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}

 

 

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