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3 years ago

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Problem #2: The radar beam sent out by an airplane reaches a distance of 120 kilometers and covers an angle of 150°. Calculate t

he area covered by the beam. 120 km 150

1 answer:

Paraphin [41]3 years ago

7 0

By the number 987 add 4 then times 7 up the cow

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6th grade math help me please ! :))

Stells [14]

Answer:

1,825 days

Step-by-step explanation:

730 / 2 = 365 (one year, or the number of days a single revolution around the sun takes)

Now just multiply by 5

[365 · 5 = 1,825], thus 1,825 days are the duration of 5 revolutions.

The problem could also be initially expressed as:

x = (720/2) · (5) with x representing the missing value

6 0

3 years ago

What is the simplified form of the following expression?

nika2105 [10]

Answer:

I believe your answer is A

Hopefully I could help :)

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4 years ago

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Is there any number that when added to 0 is negative, and when subtracted by 0 is also negative?

bogdanovich [222]

Answer:

Subtracting with zero: 0 – a = –a. Use the rule for subtracting signed numbers: Change the operation from subtraction to addition and change the sign of the second number. Likewise, a – 0 = a. It doesn’t change the value of a to subtract zero from it.

Step-by-step explanation:

6 0

3 years ago

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The attendance at baseball games at a certain stadium is normally distributed, with a mean of 30,000 and a standard deviation of

lakkis [162]

Answer:

a) 0.964

b) 0.500

c) 0.885

d) $x \geq 32467.5$

e) 0.997

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 30000

Standard Deviation, σ = 1500

We are given that the distribution of attendance at stadium is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(attendance is greater than 27,300)

P(x > 27300)

$P( x > 27300) = P( z > \displaystyle\frac{27300 - 30000}{1500}) = P(z > -1.8)$

$= 1 - P(z \leq -1.8)$

Calculation the value from standard normal z table, we have,

$P(x > 27300) = 1 - 0.036 = 0.964 = 96.4\%$

b) P(attendance greater than or equal to 30000)

$P(x > 30000) = P(z > \displaystyle\frac{30000-30000}{1500}) = P(z \geq 0)\\\\P( z \geq 0) = 1 - P(z \leq 0)$

Calculating the value from the standard normal table we have,

$1 - 0.500 = 0.500 = 50\%$

c) P(attendance between 27000 and 32000)

$P(27000 \leq x \leq 32000) = P(\displaystyle\frac{27000 - 30000}{1500} \leq z \leq \displaystyle\frac{32000-30000}{1500}) = P(-2 \leq z \leq 1.33)\\\\= P(z \leq 1.33) - P(z < -2)\\= 0.908 - 0.023 = 0.885 = 88.5\%$

$P(27000 \leq x \leq 32000) = 88.5\%$

e) P(attendance less than 33000)

P(x < 33000)

$P( x < 33000) = P( z < \displaystyle\frac{33000 - 30000}{1500}) = P(z < 2)$

Calculating the value from the standard normal table we have,

P(x < 33000) = 0.997 = 99.7%

d) We have to find x such that:

$P( X > x) = P( z > \displaystyle\frac{x - 30000}{1500}) = 0.95$

Calculating the value from the standard normal table we have,

P(z = 1.645) = 0.95

Thus,

$\displaystyle\frac{x - 30000}{1500} \geq 1.645\\\\x \geq 32467.5$

The attendance should be greater than or equal to 32467.5 to be in the top 5% of all games.

6 0

4 years ago

Tanner has already taken 1 picture at home, and he expects to take 2 pictures during every day of vacation. How many days will T

SIZIF [17.4K]

Answer:

38 is the answer

Step-by-step explanation:

2 times 19 is 38

4 0

4 years ago

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