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3 years ago

In a class of 666, there are 222 students who forgot their lunch.

Mathematics

2 answers:

pychu [463]3 years ago

8 0

Is the numbers just 6, 2 and 2 if those are the numbers the answer is 1/15

pav-90 [236]3 years ago

5 0

Answer:

1/15

Step-by-step explanation:

The probability that the first student has forgotten their lunch is 2/6.

Since there are only 5 students left the probability that the second student that is picked and has forgotten their lunch is 1/5.

$2/6*1/5=2/30=1/15\\$

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Uhhhhhhhhhh I think its is 84

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3 years ago

A rectangle has a perimeter of 165.4 kilometers and a height of 20 kilometers. What is the length of the base?

KATRIN_1 [288]

Answer: Follow the Steps Below.

Step-by-step explanation: Simply divide the area of the rectangle by its height to find its base. Other forms of solving for the base can be accomplished knowing diagonal length by simply taking the square root of the diagonal length squared minus its height squared.

5 0

3 years ago

Ryan and Sarah collect baseball cards. Ryan has 150 baseball cards and collects 10 cards per week. Sarah has 200 baseball cards

Nadusha1986 [10]

Answer:

<h2>10 weeks </h2>

Step-by-step explanation:

Step one:

given data

Ryan

number of baseball cards=150

number collected per week= 10

let the number of weeks be x

and the total be y

y=10x+150-----------------1

Sarah

number of baseball cards=200

number collected per week= 5

let the number of weeks be x

and the total be y

y=5x+200------------2

Step two:

Required

the number of weeks where both total will be the same

10x+150=5x+200

10x-5x=200-150

5x=50

divide both sides by 5

x=50/5

x=10 weeks

3 0

3 years ago

2. When a large truckload of mangoes arrives at a packing plant, a random sample of 150 is selected and examined for

kirza4 [7]

a) The 90% confidence interval of the percentage of all mangoes on the truck that fail to meet the standards is: (7.55%, 12.45%).

b) The margin of error is: 2.45%.

c) The 90% confidence is the level of confidence that the true population percentage is in the interval.

d) The needed sample size is: 271.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions has the bounds given by the rule presented as follows:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

The variables are listed as follows:

$\pi$ is the sample proportion, which is also the estimate of the parameter.

z is the critical value.

n is the sample size.

The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of $\frac{1+0.90}{2} = 0.95$ , so the critical value is z = 1.645.

The sample size and the estimate are given as follows:

$n = 150, \pi = \frac{15}{150} = 0.1$

The margin of error is of:

$M = z\sqrt{\frac{0.1(0.9)}{150}} = 0.0245 = 2.45\%$

The interval is given by the estimate plus/minus the margin of error, hence:

The lower bound is: 10 - 2.45 = 7.55%.

The upper bound is: 10 + 2.45 = 12.45%.

For a margin of error of 3% = 0.03, the needed sample size is obtained as follows:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.1(0.9)}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.1(0.9)}$

$\sqrt{n} = \frac{1.645\sqrt{0.1(0.9)}}{0.03}$

$(\sqrt{n}})^2 = \left(\frac{1.645\sqrt{0.1(0.9)}}{0.03}\right)^2$

n = 271 (rounded up).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

3 0

2 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

<u>Prove that:</u>

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

<u>Proof: </u>

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

<u>Taking</u><u> </u><u>LHS</u>

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

<em>On combining the fractions,</em>

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

<em>Regrouping the terms,</em>

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

3 years ago