Below is the number of games each quarterback in the Northwest division has played this season.

Which two triangles are congruent? Complete the congruence statement.

andre [41]

answer:

△JKI ≅ △CED

step-by-step explanation:

like i mentioned before, they look similar and fulfill the SSS triangle theorem (this time's the theorem is different only)

4 0

2 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

2 years ago

John has 5,000 to invest in a saving account that has interest compounded annually. If he wants his money to double in 8 years w

ololo11 [35]

hi

5 000 * X ^(8) = 10 000

x^(8) = 10 000 / 5000

x ^(8)= 2

x = 2^(1/8)

x≈ 1.09

intrest rate must be 9%

7 0

2 years ago

Read 2 more answers

If the total income generated from gasoline for AER was $408 million how much would be the cost for a barrel of gasoline, gasoli

Mazyrski [523]

The problem statement appears to be trying to tell you that 60 million barrels of crude were processed, resulting in 34% of that volume being turned into gasoline, which was then sold for a total of $408 million. You are asked for the revenue associated with 1 barrel of gasoline.

($408·10^6)/(60·10^6 bbl × 0.34) = $408/(20.4 bbl) = $20/bbl

The income from one barrel of gasoline is $20.00.

3 0

2 years ago

Read 2 more answers

What is (c+1) +9 in simplified form plz help

Afina-wow [57]

=(c+1)+9
=C+1+9
=C+10
10c

7 0

3 years ago

Read 2 more answers