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2 years ago

14

CAN SOMEONE PLZzzz Plzzzz HELP!!!???

1 answer:

umka21 [38]2 years ago

8 0

First plot the points. Let’s just use the first graph. When you have done that. Draw a triangle. Find the right angle and look at what line is across from that. That is the hypotenuse. That is the length that you are trying to find. So you have to do your equation: a^2 + b^2 = c^2. A and B have to be the length of the other 2 lines(just count it). When you have done that repack you a and b with your #s. And what ever is equal to you c. Then that is you answer. (Im sorry if this was confusing)

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The expression m÷4 is the distance each person runs in a relay race that is m miles. How far does each person run in a relay rac

inna [77]

Answer:

5

Step-by-step explanation:

Replace 20 instead of m

20/4=5

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2 years ago

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PLS HELPP PLSS this homework is so hard

Illusion [34]

So you should label the x axis and y axis like I did then you use the formula!

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1 year ago

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ABC Auto Insurance classifies drivers as good, medium, or poor risks. Drivers who apply to them for insurance fall into these th

Misha Larkins [42]

Answer:

a. $P(E_1/A)=0.0789$

b. $P(E_2/A)=0.395$ \

c. $P(E_3/A)=0.526$

Step-by-step explanation:

Let $E_1,E_2,E_3$ are the events that denotes the good drive, medium drive and poor risk driver.

$P(E_1)=0.30,P(E_2)=0.50,P(E_3)=0.20$

Let A be the event that denotes an accident.

$P(A/E_1)=0.01$

$P(A/E_2=0.03$

$P(A/E_3)=0.10$

The company sells Mr. Brophyan insurance policy and he has an accident.

a.We have to find the probability Mr.Brophy is a good driver

Bayes theorem, $P(E_i/A)=\frac{P(A/E_i)\cdot P(E_1)}{\sum_{i=1}^{i=n}P(A/E_i)\cdot P(E_i)}$

We have to find $P(E_1/A)$

Using the Bayes theorem

$P(E_1/A)=\frac{P(A/E_1)\cdot P(E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}$

Substitute the values then we get

$P(E_1/A)=\frac{0.30\times 0.01}{0.01\times 0.30+0.50\times 0.03+0.20\times 0.10}$

$P(E_1/A)=0.0789$

b.We have to find the probability Mr.Brophy is a medium driver

$P(E_2/A)=\frac{0.03\times 0.50}{0.038}=0.395$

c.We have to find the probability Mr.Brophy is a poor driver

$P(E_3/A)=\frac{0.20\times 0.10}{0.038}=0.526$

7 0

3 years ago

Would appreciate some help with this:<br><br> Verify (sinq+cosq)^2 - 1 = sin2q

Phantasy [73]

Hello from MrBillDoesMath!

Answer:

See Discussion below

Discussion:

(sinq + cosq)^2 = => (a +b)^2 = a^2 + 2ab + b^2

(sinq)^2 + (cosq)^2 + 2 sinq* cosq => as (sinx)^2 + (cosx)^2 = 1

1 + 2 sinq*cosq (*)

Setting a = b = q in the trig identity:

sin(a+b) = sina*cosb + cosa*sinb

sin(2q) = (**)

sinq*cosq + cosq*sinq => as both terms are identical

2 sinq*cosq

Combining (*) and (**)

(sinq + cosq)^2 = 1 + 2sinq*cosq => (**) 2sinq*cosq = sqin(2q)

= 1 + sin(2q)

Hence

(sinq + cosq)^2 = 1 + sin(2q) => subtracting 1 from both sides

(sinq + cosq)^2 - 1 = sin(2q)

The last statement is what we are trying to prove.

Thank you,

MrB

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2 years ago

3xxyyy using exponents

olasank [31]

Just count the amount of each letter and write it as a exponent = 3x²y³

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2 years ago

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