if the texting portion of your plan for $15 a month and you said 500 text messages in a month how much does it cost per text

What is the length of BC¯¯¯¯¯ ?

balandron [24]

Answer:

The length of BC is 17

Step-by-step explanation:

When we construct a triangle with the data, we will end up a isosceles triangle.

In a isosceles triangle , we know that tow angles and two sides are equal

The questions states that

$\angle b = \angle c$

Then sides AC and AB are also equal

On equation their values

2X - 8 = X+9

2X = X + 9 +8

2X = X + 17

2X - X = 17

X = 17

The length of BC = X = 17

4 0

3 years ago

Need help & Thank you !!!!

loris [4]

The right answer is of option D.

please see the attached picture for full solution...

Hope it helps...

Good luck on your assignment......

8 0

3 years ago

Al, Tom and Joe share €3000.

agasfer [191]

Answer:

Tom's share = €800

Step-by-step explanation:

Given that:

Amount shared by Al, Tom and Joe = €3000

Ratio of Al's share to Tom's share = 5 : 4

Joe's share = 1.5(Tom's share) = 1.5(4) = 6

Therefore,

Ratio of Al, Tom and Joe = 5 : 4 : 6

Let,

x be the number of times of amount each get.

5x + 4x + 6x = 3000

15x = 3000

Dividing both sides by 15

$\frac{15x}{15}=\frac{3000}{15}\\x=200$

Tom's share = 4x = 4(200) = €800

Hence,

Tom's share = €800

5 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:

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

Thus, the value of P (X > 20) is 0.1353.

(c)

Compute the value of P (X < 30) as follows:

$P(X$

Thus, the value of P (X < 30) is 0.9502.

(d)

It is given that, P (X < x) = 0.95.

Compute the value of <em>x</em> as follows:

$P(X$

Take natural log on both sides.

$ln(e^{-0.10x})=ln(0.05)\\-0.10x=-2.996\\x=\frac{2.996}{0.10}\\ =29.96\\\approx30$

Thus, the value of x is 30.

7 0

3 years ago

During the first part of a trip, a canoeist travels 18 miles at a certain speed. the canoeist travels 4 miles on the second par

storchak [24]

We can set it up like this, where <em>s </em>is the speed of the canoeist:

$\frac{18}{s} + \frac{4}{s-5} = 3$

To make a common denominator between the fractions, we can multiply the whole equation by s(s-5):

$s(s-5)[\frac{18}{s} + \frac{4}{s-5} = 3] \\ 18(s-5)+4s=3s(s-5) \\ 18s - 90+4s=3 s^{2} -15s$

If we rearrange this, we can turn it into a quadratic equation and factor:

$18s - 90+4s=3 s^{2} -15s \\ 22s-90=3 s^{2} -15s \\ 3 s^{2} -37s+90=0 \\ (3s-10)(s-9)=0 \\ s= \frac{10}{3} ,9$

Technically, either of these solutions would work when plugged into the original equation, but I would use the second solution because it's a little "neater." We have the speed for the first part of the trip (9 mph); now we just need to subtract 5mph to get the speed for the second part of the trip.

$9-5 = 4$

The canoeist's speed on the first part of the trip was 9mph, and their speed on the second part was 4mph.

5 0

3 years ago