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3 years ago

Can someone explain this?

Mathematics

2 answers:

natulia [17]3 years ago

8 0

Answer:

58 degrees

Step-by-step explanation:

(x+45)+(x+66)=95

2x+45+66=95

2x+111=95

2x=95-111

2x=-16

x=-16/2

x=-8

x+66=-8+66=58

yan [13]3 years ago

5 0

<h3>Answer: angle JML is 58 degrees (choice C)</h3>

=======================================

Explanation:

Check out the attached image below. I have marked the angles with A, B, and C such that

A = angle NMJ = x+45 (in red)

B = angle JML = x+66 (in blue)

C = angle NML = 95 (in green)

The two angles A and B combine to form angle C. Angles A and B are adjacent angles.

So,

A+B = C

(x+45) + (x+66) = 95

2x+111 = 95

2x+111-111 = 95-111 .... subtract 111 from both sides

2x = -16

2x/2 = -16/2 .... divide both sides by 2

x = -8

From here we can find the angle measures

A = angle NMJ = x+45 = -8+45 = 37 degrees

B = angle JML = x+66 = -8+66 = 58 degrees

As a check,

A+B = 37+58 = 95 = C

so the answer is confirmed.

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photoshop1234 [79]

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

There are 27 girls and 36 boys who want to participate in a math competition. If each team must have the same ratio of girls and

Reptile [31]

Since the ratio of girls and boys must be the same in each team, and all the competitors must be involved, it means that each team's ratio of boys : girls must be equal to the total ratio of Boys : Girls = 36 :27= 4 / 3.

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2 years ago

Evaluate the expression. 4.2°

Mariana [72]

Answer:

4.2°

Step-by-step explanation:

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I hope this helps!

4 0

2 years ago

A gambler has a coin which is either fair (equal probability heads or tails) or is biased with a probability of heads equal to 0

yawa3891 [41]

Answer:

(a) 0.1719

(b) 0.3504

Step-by-step explanation:

For every coin the number of heads follows a Binomial distribution and the probability that x of the 10 times are heads is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{10-x}$

Where n is 10 and p is the probability to get head. it means that p is equal to 0.5 for the fair coin and 0.3 for the biased coin

So, for the fair coin, the probability that the number of heads is less than 4 is:

$P(x$

Where, for example, P(0) and P(1) are calculated as:

$P(0)=\frac{10!}{0!(10-0)!}*0.5^0*(1-0.5)^{10-0}=0.0009\\P(1)=\frac{10!}{1!(10-1)!}*0.5^1*(1-0.5)^{10-1}=0.0098$

Then, $P(x$ , so there is a probability of 0.1719 that you conclude that the coin is biased given that the coin is fair.

At the same way, for the biased coin, the probability that the number of heads is at least 4 is:

$P(x\geq4 )=P(4)+P(5)+P(6)+...+P(10)$

Where, for example, P(4) is calculated as:

$P(4)=\frac{10!}{4!(10-4)!}*0.3^4*(1-0.3)^{10-4}=0.2001$

Then, $P(x\geq4 )=0.3504$ , so there is a probability of 0.3504 that you conclude that the coin is fair given that the coin is biased.

7 0

3 years ago