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Westkost [7]
3 years ago
7

Helppp please. I’ve been stuck for days

Mathematics
2 answers:
GalinKa [24]3 years ago
6 0
The answer for this question is 11
m_a_m_a [10]3 years ago
3 0

Answer:x=11

Step-by-step explanation:5 x 11= 55 + 2 = 57 and 3 x 11 = 33 and 57 + 33 = 90

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Solve this to get 25 points. Correct answer gets brainliest!
Fantom [35]

Answer:

T = 59

U = 27

V = 94

Step-by-step explanation:

(5x +4) + (8x + 6) + (2x+5) = 180

15x + 15 = 180

15x = 165

x = 11

7 0
3 years ago
Find the length of the hypotenuse in the figure shown below.
slega [8]

Answer:

the answer is b.13 meters

Step-by-step explanation:

a²+b²=c²

3 0
3 years ago
Read 2 more answers
The area of the shaded sector is shown. Find the radius of circle M. Round your answer to the nearest hundredth.
Vaselesa [24]

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete, as the sector, its area and the angle at the center are not given.

I will solve this using the following illusration

The area of a sector is:

Area= \frac{\theta}{360} * \pi r^2

Assume that:

\theta = 120

Area = 20

The equation becomes

20= \frac{120}{360} * \pi r^2

Simplify

20= \frac{1}{3} * \pi r^2

Take: \pi = 3.14

20= \frac{1}{3} *3.14* r^2

Cross Multiply

3.14 * r^2 = 20 * 3\\

Solve for r^2

r^2 = \frac{20 * 3}{3.14}

r^2 = 19.11

Take square roots

r = \sqrt{19.11

r = 4.4

3 0
2 years ago
Maya is 14 years old. Her brother Jorge is 3 years more than half her age. Which of the following is the correct expression for
Elina [12.6K]

Answer:

10

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger
11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let <em>X</em> = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = p=\frac{1}{120}.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X </em>is:

P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...

In this case we need to compute the probability of 1 or more than 1 catastrophes in <em>n</em> missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             =1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419

6 0
3 years ago
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