HELP What is the length of FC?

Jarvis works in a garage £6 an hour. If he works on Saturday, he is paid time and a quarter. If he works on Sunday, he is paid t

xxMikexx [17]

Answer: He is paid $90 last weekend.

Step-by-step explanation:

Since we have given that

Amount he earns per hour = $6

If he works on Saturday, he is paid time and a quarter .

Amount would be

$6\times 1.25=\$7.5$

If he works on Sunday, he is paid time and a half.

Amount would be

$6\times 1.5=\$9$

Number of hours he worked on Saturday = 6 hours

Number of hours he worked on Sunday = 5 hours

So, Total amount he is paid last weekend altogether is given by

$6\times 7.5+5\times 9\\\\=\$90$

Hence, he is paid $90 last weekend.

5 0

3 years ago

What is the length of an arc with a central angle of 23π radians and a radius of 24 cm?

dalvyx [7]

We can solve for the arc length using the formula shown below:
Arc Lenght = 2pi*r(central angle/360)

We need to convert the central angle such as:
Central angle = 23pi rad * (180°/pi rad) = 23*180
Radius = 24cm

Solving for arc length:
Arc length = 2*3.14 *24*(23*180/360)
Arc length = 1733.28 cm

7 0

3 years ago

What is the probability that a random person who tests positive for a certain blood disease actually has the disease, if we know

xeze [42]

Answer:

Step-by-step explanation:

Hello!

Any medical test used to detect certain sicknesses have several probabilities associated with their results.

Positive (test is +) ⇒ P(+)

True positive (test is + and the patient is sick) ⇒ P(+ ∩ S)

False-positive (test is + but the patient is healthy) ⇒P(+ ∩ H)

Negative (test is -) ⇒ P(-)

True negative (test is - and the patient is healthy) ⇒ P(- ∩ H)

False-negative (test is - but the patient is sick) ⇒ P(- ∩ S)

The sensibility of the test is defined as the capacity of the test to detect the sickness in sick patients (true positive rate).

⇒ P(+/S) = P(+ ∩ S)

P(S)

The specificity of the test is the capacity of the test to have a negative result when the patients are truly healthy (true negative rate)

⇒ P(-/H) = P(- ∩ H)

P(H)

For this particular blood disease the following probabilities are known:

1% of the population has the disease: P(S)= 0.01

95% of those who are sick, test positive for it: P(+/S)= 0.95 (sensibility of the test)

2% of those who don't have the disease, test positive for it: P(+/H)= 0.02

The probability of a person having the blood sickness given that the test was positive is:

P(S/+)= P(+ ∩ S)

P(+)

The first step you need to calculate the intersection between both events + and S, for that you will use the information about the sickness prevalence in the population and the sensibility of the test:

P(+/S) = P(+ ∩ S)

P(S)

P(+/S)* P(S) = P(+ ∩ S)

P(+ ∩ S) = 0.95*0.01= 0.0095

The second step is to calculate the probability of the test being positive:

P(+)= P(+ ∩ S) + P(+ ∩ H)

Now we know that 1% of the population has the blood sickness, wich means that 99% of the population doesn't have it, symbolically: P(H)= 0.99

Then you can clear the value of P(+ ∩ H):

P(+/H) = P(+ ∩ H)

P(H)

P(+/H)*P(H) = P(+ ∩ H)

P(+ ∩ H) = 0.02*0.99= 0.0198

Next you can calculate P(+):

P(+)= P(+ ∩ S) + P(+ ∩ H)= 0.0095 + 0.0198= 0.0293

Now you can calculate the asked probability:

P(S/+)= P(+ ∩ S) = 0.0095 = 0.32

P(+) 0.0293

I hope it helps!

6 0

2 years ago

In the figure above, the vertices of square

Olin [163]

(C) 6 + 3√3

Explanation:

Area of the square = 3

a X a = 3

a² = 3

a = √3

Therefore, QR, RS, SP, PQ = √3

ΔBAC ≅ ΔBQR

Therefore,

$\frac{BQ}{BA} = \frac{QR}{AC}$

$\frac{BQ}{BA} = \frac{\sqrt{3} }{BA}$

In ΔBAC, BA = AC = BC because the triangle is equilateral

So,

BQ = √3

So, BQ, QR, BR = √3 (equilateral triangle)

Let AP and SC be a

So, AQ and RC will be 2a

In ΔAPQ,

(AP)² + (QP)² = (AQ)²

(a)² + (√3)² = (2a)²

a² + 3 = 4a²

3 = 3a²

a = 1

Similarly, in ΔRSC

(SC)² + (RS)² = (RC)²

(a)² + (√3)² = (2a)²

a² + 3 = 4a²

3 = 3a²

a = 1

So, AP and SC = 1

and AQ and RC = 2 X 1 = 2

Therefore, perimeter of the triangle = BQ + QA + AP + PS + SC + RC + BR

Perimeter = √3 + 2 + 1 + √3 + 1 + 2 + √3

Perimeter = 6 + 3√3

Therefore, the perimeter of the triangle is 6 + 3√3

8 0

3 years ago

How do I solve cos x> 1/2 sin x?

Hatshy [7]

Divide both sides by $\dfrac12\cos x$ :

$\dfrac{\cos x}{\frac12\cos x}>\dfrac{\frac12\sin x}{\frac12\cos x}\implies2>\dfrac{\sin x}{\cos x}=\tan x$

If $-\dfrac\pi2$ , then

$\tan x$

and more generally, for any integer $n$ ,

$n\pi-\dfrac\pi2$

4 0

2 years ago