I cannot do graphs to save my life someone help!

What is the value of x? The figure contains 2 lines that intersect to form what looks like an x. Two of the 4 angles in the x ar

Gnom [1K]

Answer:

$x=38.5^\circ$

Step-by-step explanation:

Given that the left angle = $(3x-3)^\circ$

The right angle across from it $= 6(x-10)^\circ$

The other two angles are x and x.

We know that the sum of angles at a point equals 360 degrees.

Therefore,

$(3x-3)^\circ+6(x-10)^\circ+x+x=360^\circ\\3x-3+6x-60+2x=360\\11x-63=360\\11x=360+63\\11x=423\\x=38.5^\circ$

The value of x is approximately 38.5 degrees.

6 0

3 years ago

Joey works 20 hours a week. If he works for 10 months continually, how many hours will he have worked? (1 month = 4 weeks)

Tcecarenko [31]

800 because 20*4*10=800 which means he works 80 weeks a month for 10 months

3 0

3 years ago

In a bag of 18 oranges, 12 have gone bad.<br><br> What is the ratio of good oranges to bad oranges?

klemol [59]

Of 18 oranges, if 12 are bad that means 6 are good. Ratio of good to bad is 6:12 reduced to 1:2, the answer

4 0

3 years ago

Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X > 10) b. P(X > 20) c

GrogVix [38]

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(c) The value of P (X < 30) is 0.9502.

(d) The value of x is 30.

Step-by-step explanation:

The probability density function of an exponential distribution is:

$f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0$

The value of E (X) is 10.

The parameter λ is:

$\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10$

(a)

Compute the value of P (X > 10) as follows:

$P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679$

Thus, the value of P (X > 10) is 0.3679.

(b)

Compute the value of P (X > 20) as follows:

$P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353$