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3 years ago

PLEASE HELP. use the information provided to write the equation of each circle . i will give brainliest!

Mathematics

1 answer:

igor_vitrenko [27]3 years ago

4 0

Answer: 16. C 17. A 18. A

19 You got this! There is only one answer with the correct radius squared!

Step-by-step explanation:

For the coordinates of the center, flip the signs in the parentheses.

For the "=output": in the equation, square the given radius.

Center: (-14.4) the signs in (x + 14)²+(y - 4)² are the opposites.

Radius given: 2 Square it to get the =4

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Galina-37 [17]

Y=12
x=6

The triangle is a 30-60-90 triangle!

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3 years ago

Help me plz help assao

8_murik_8 [283]

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3 years ago

The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest

Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

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

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

$P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)$

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

$P(x < 5) =0.7287= 72.87\%$

b) P( port handles 3 or more million tons of cargo per week)

$P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)$

Calculating the value from the standard normal table we have,

$1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%$

c)P( port handles between 3 million and 4 million tons of cargo per week)

$P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%$

$P(3 \leq x \leq 4) = 23.7\%$

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85$

$= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85$

$=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15$

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

$P( z \leq -1.036) = 0.15$

$\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65$

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

8 0

3 years ago

if two angles are complementary then the sum of their measure is 90. if the sum of the measures of two angles is 90 then both of

Tju [1.3M]

Yes, both angles are acute because all complementary angles are acute.

6 0

2 years ago

Hi, i need to know the work and answer to this. thanks!

Nesterboy [21]

Answer:

432 in^2

Step-by-step explanation:

in similar quadrilaterals, the first point of one quad. corresponds to the first point of the other quad, so in this case UA corresponds with CH.

since CH is 3/4 the length of UA, we can also assume that the other sides in ZUCH are 3/4 the length of their corresponding sides in SQUA.

even though we don't know what quadrilateral SQUA and ZUCH are, we know the area of SQUA is 9/16 times less than ZUCH.

want some proof?

lets say SQUA and ZUCH are rectangle/square

ZUCH: 4X4 = 16

SQUA: 3X3 = 9

now lets say they are trapezoids. We will set ZUCH 2nd base to 8 and height to 16, therefore SQUA bases will be 3 and 6, and the height will be 12 (multiply ZUCH lengths by 3/4)

ZUCH = (b1+b2)(h)/2 = (4+8)(16)/2 = 96

SQUA = (b1+b2)(h)/2 = (3+6)(12)/2 = 54

simplify 96/54 = 16/9

now we can multiply 243 by our factor 16/9 to find the area of SQUA.

243 * 16/9 = 432 in^2

6 0

3 years ago