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pashok25 [27]
3 years ago
14

Help please need those answers

Mathematics
1 answer:
icang [17]3 years ago
8 0

We know: The sum of the measures of the angles of a triangle is equal 180°.

We have: m∠A =65°, m∠B =  (3x - 10)° and m∠C = (2x)°.

The equation:

65 + (3x - 10) + 2x = 180

(3x + 2x) + (65 - 10) = 180

5x + 55 = 180        <em>subtract 55 from both sides</em>

5x = 125          <em>divide both sides by 5</em>

x = 25

m∠B =  (3x - 10)° → m∠B =  (3 · 25 - 10)° = (75 - 10)° = 65°

m∠C = (2x)° → m∠C = (2 · 25)° = 50°

<h3>Answer: x = 25, m∠B = 65°, m∠C = 50°</h3>
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A newspaper conducted a survey to find out how many high school students play video games.The two ways table below displays the
zepelin [54]

Answer:

Step-by-step explanation:

Given that

We are given the following:

1. Boys who do play video games = 1593

2. Boys who do not play video games = 858

3. Total boys = 2451

4. Girls who do play video games = 1361

5. Boys who do not play video games = 1635

6. Total Girls = 2996

7. Total number of candidates = Total boys + Total girls

=(2451 + 2996) = 5447

8. Total number of students, who play video games = 2954

9. Total number of students, who do not play video games = 2493

Percentage of boys who play video games:

\dfrac{1593}{2451} \times 100 = 64.99\%

So, option A is not correct.

Percentage of girls who play video games:

\dfrac{1361}{2996} \times 100 = 45.43\%

so option B is correct. The percentage is near about 45%,

Percentage of Total number of students, who do not play video games:

\dfrac{2493}{5447} \times 100 = 45.77\%

So, option C is not correct.

7 0
3 years ago
A page in Johanie’s workbook is torn, and one of the questions is cut off. She can read only the first part of the question:
kirill115 [55]
B quadrant ll or quadrant lll
5 0
3 years ago
Your state charges 5% sales tax on nonfood items. You're buying a light bulb for $3.00, cereal for
dimulka [17.4K]
We first add the tax to the non-food items.  We have ($3.00+$.50)*1.05=($3.50)*1.05=$3.675 (Which can be rounded up to $3.68, since we can't have a fraction of a cent).

Now, we add the cost of the food to $3.68.  As a result, we have $3.68+$2.50=$6.18 as the total bill.
6 0
3 years ago
How do you solve this? 18 " n=21
dalvyx [7]
18 ^{n} =21

log(18) ^{n}=log(21)

n*log(18)=log(21)

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6 0
3 years ago
A store selling newspapers orders only n = 4 of a certain newspaper because the manager does not get many calls for that publica
umka2103 [35]

Answer:

a) The expected value is 2.680642

b) The minimun number of newspapers the manager should order is 6.

Step-by-step explanation:

a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be <em>more</em> than that.

X is a random variable of Poisson distribution with mean \mu = 3 , and Y is a random variable with range {0, 1, 2, 3, 4}, with the following values

  • PY(k) = PX(k) = ε^(-3)*(3^k)/k! for k in {0,1,2,3}
  • PY(4) = 1 -PX(0) - PX(1) - PX(2) - PX (3)

we obtain:

PY(0) = ε^(-3) = 0.04978..

PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936

PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404

PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404

PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768

E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) =  0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642

The store is <em>expected</em> to sell 2.680642 newspapers

b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05

we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231

FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262

FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608

FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489

So, if we ask for 6 newspapers, the probability of receiving at least 6 calls is 0.966489, and the probability to receive more calls than available newspapers will be less than 0.05.

I hope this helped you!

8 0
3 years ago
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