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3 years ago

Get 14 points, Plz help me with this question, and give the right answer cause it's important

Mathematics

2 answers:

garri49 [273]3 years ago

6 0

Answer:

it's 20

Step-by-step explanation:

(X+5)° + (7x-5)° + x°= 180° { sum of the three angles of the triangle is 180°}

X+5+7x-5+X = 180

X+7x+X+5-5= 180

9x+0 = 180

9x = 180

therefore X= 180/9

= 20

Nitella [24]3 years ago

5 0

All angles are of triangle and thus give sum of 180⁰

x + 5 + 7x - 5 + x = 180⁰

9x = 180⁰

x = 20⁰

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Answer:

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3 years ago

The height of a cone-shaped statue is 9 ft, and the diameter is 12 ft.

Pepsi [2]

I hope this helps you

Volume of cone =1/3.pi.r^2.h

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3 0

3 years ago

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Solve by finding the square root 2x^2=32

Cerrena [4.2K]

2x^2=32 so divide by 2 to make x^2=16 and the square root so x=4

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3 years ago

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100 points

STatiana [176]

Answer:

The distance traveled by the B737-400 aircraft in 5 hours is 2,000 miles

Step-by-step explanation:

Average speed = 400 miles per hour

Distance = 400t

t= 5

Total distance = 400t at t hours

Total distance when t = 5 hours

Total distance traveled = 400t

= 400 × 5

= 2,000 miles

Total distance travelled = 2,000 miles

The distance traveled by the B737-400 aircraft in 5 hours is 2,000 miles

hope this helps

7 0

3 years ago

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2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$

The z-score corresponding to 4.28 is 0.99999

$P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\$

So it means that it is very unlikely that there will be more than 980 non-defective units.

8 0

4 years ago