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3 years ago

SOMEONE HELP ME DO THESE 4 PROBLEMS FOR 50 points and BRAINLIEST.

Mathematics

2 answers:

USPshnik [31]3 years ago

6 0

Answer:

1. hypotenuse is 25

2. hypotenuse is 7.8

3. hypotenuse is 9.8

4. hypotenuse is 3.6

Step-by-step explanation:

To find the hypotenuse, just use the formula a^2 + b^2 = c^2

a and b are the sides with short lengths, while c is the one with the greatest line.

Let's try #1, and then you can do the rest by yourself, as it is very easy.

1) sides a and b are given, a = 7 and b = 24

hypotenuse = a^2 + b^2 = c^2

--> 7a^2 + 24b^2 = c^2

--> 49 + 576 = c^2

--> 625 = c^2

--> take the square root of 625

--> 25 = c

Norma-Jean [14]3 years ago

3 0

Answer:

Step-by-step explanation:

2. 24^2+7^2 = 625

V 625 = 25

3. 6^2 + 5^2 = 25 +36 = 61

V 61 =7.81024967 =7.8

4. 4^2 +9^2 = 97

V 97= 9.84885780 = 9.8

5. 2^2 +3^2 = 13

V 13 =3.60555127 =3.6

and done!

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Solve the problem. A $8920 note is signed, for 160 days, at a discount rate of 6.5 % Find the proceeds.

GuDViN [60]

<h2>Hello!</h2>

The answer is $101688

Since there is not enough information about the rate, let's assume that it's a daily rate.

<h2>Calculations</h2>

Let's say:

P : Proceeds

o : original amount

d : number of days

r : percentual rate

So,

$P = o + o*(r*d)\\$

Let's make the percentual rate (6,5%) a real number by dividing into 100,

$\frac{6,5}{100}=0,065$

By substituting, we have

$P = $8920 + $8920*(0,065*160)= 8920 + 8920*10,4\\P= $8920 + $92768 = $101688$

$P =$ $101688

Have a nice day!

5 0

4 years ago

Read 2 more answers

Find the first three term if the general nth term is T = 2-n

Elena L [17]

Tn = 2n - 6

to find first term ,n= 1

substitute n for 1 in the equation

T1 = 2(1) - 6

T1 = 2 - 6 =-4

to find second term , n =2

substitute n for 2 in the equation

T2 = 2(2) - 6

T2 = 4 - 6 = -2

to find third term, n = 3

substitute n for 3 in the equation

T3 = 2(3) - 6

T3 = 6–6 = 0

FIRST TERM = -4

SECOND TERM = -2

THIRD TERM = 0

8 0

3 years ago

Solve 3x+2/2=13<br>really need help fast ;-;

kirza4 [7]

Answer:

3x+2=26

3x=24

x=8

Step-by-step explanation:

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3 years ago

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Solving Equations In Exercise,solve the equation for<br> e^(x)1/2 = e^3

ra1l [238]

Answer:

x = 9

Step-by-step explanation:

We are given the equation

$e^{(x)^{1/2}} = e^3$

We solve the equation in the following manner

$e^{(x)^{1/2}} = e^3\\e^{\sqrt{x}} = e^3\\\text{Comparing both the sides we get}\\\sqrt{x} = 3\\\text{Squaring both sides}\\(\sqrt{x})^2 = 3^2\\x = 9$

Solving for x, we get

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8 0

3 years ago

The sensitivity is about 0.993. That is, if someone has HIV, there is a probability of 0.993 that they will test positive. • The

seraphim [82]

Answer:

(a) The probability that someone will test positive and have HIV is 0.000025.

(b) The probability that someone will test positive and not have HIV is 0.0001.

(d) The probability a person has HIV given that he/she was tested positive is 0.1986.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person has HIV

<em>Y</em> = a person is tested positive for HIV.

The information provided is:

$P(Y|X)=0.993\\P(Y^{c}|X^{c})=0.9999\\P(X)=0.000025$

Compute the probability of a person not having HIV as follows:

$P(X^{c})=1-P(X)=1-0.000025=0.999975$

Compute the probability of $(Y^{c}|X)$ as follows:

$P(Y^{c}|X)=1-P(Y|X)=1-0.993=0.007$

Compute the probability of as $(Y|X^{c})$ follows:

$P(Y|X^{c})=1-P(Y^{c}|X^{c})=1-0.9999=0.0001$

(a)

Compute the probability that someone will test positive and have HIV as follows:

$P(Y\cap X)=P(Y|X)P(X)\\=0.993\times0.000025\\=0.000024825\\\approx0.000025$

Thus, the probability that someone will test positive and have HIV is 0.000025.

(b)

Compute the probability that someone will test positive and not have HIV as follows:

$P(Y\cap X^{c})=P(Y|X^{c})P(X^{c})\\=0.0001\times0.999975\\=0.0000999975\\\approx0.0001$

Thus, the probability that someone will test positive and not have HIV is 0.0001.

(c)

Compute the probability that someone will test positive as follows:

$P(Y)=P(Y\cap X)+P(Y\cap X^{c})=0.000025+0.0001=0.000125$

Thus, the probability that someone will test positive is 0.000125.

(d)

Compute the probability a person has HIV given that he/she was tested positive as follows:

$P(X|Y)=\frac{P(Y|X)P(X)}{P(Y)} \\=\frac{0.993\times0.000025}{0.000125}\\ =0.1986$

Thus, the probability a person has HIV given that he/she was tested positive is 0.1986.

As the probability of a person having HIV given that he was tested positive is not very large, it would not be wise to implement a random testing policy.

3 0

4 years ago